What's the best strategy to fully fill the fridge with beer bottles and have them all cooled? I'm having a party.
Suppose I'd like to have a fridge full of cold ($6~^\circ\text{C}$ or below) beer bottles, in as short a time frame as possible. The fridge indicates that it is targeting (and presumably currently at) $4~^\circ\text{C}$. All the bottles are currently at $30~^\circ\text{C}$, which is the temperature outside the fridge, and together the bottles will fill the fridge completely (as in: no further bottles could be fitted in).
What is the best strategy to achieve this aim?
(Should I put in all bottles at once? Should I put the bottles in at different times? And, if so, should I make them touch, or keep them initially apart as much as possible? Should I even consider taking out some bottles at some time and putting them back in later? (That would be really weird.) If so, should I have those bottles cool other bottles that weren't in the fridge yet?)

NB 1: I do not have a freezer available.
NB 2: Assume that I know how to get as much bottles into the fridge as possible (without breaking any).

Measurements: 1 bottle of beer: $0.61~\text{kg}$, 1 empty bottle (with cap): $0.28~\text{kg}$, 1 empty bottle (without cap): $0.28~\text{kg}$, contents of 1 bottle: $0.33~\text{l}$. My guesstimate is that I can get 72 bottles into the fridge.
 A: Without doing the analysis, I would think that a cooling system
is more effective at extracting heat from a warm container than from a
cold one.
For a fridge, the effectiveness (or coefficient of performance) is
$Eff=Q_c/W$ is the ratio of the heat removed from the cold source (the fridge) to the energy used for
the purpose. It increases with the temperature of the cold source. This is actually the prime factor to be considered in the analysis.
If we assume that, for a given current temperature of its contents, and thus for a given coefficient of performance, the cooling capacity of the fridge (heat removed per second) is limited only by the
power of its cooling engine (I do not know whether that is the case),
then the heat pump will pump more heat per second when the fridge is warm.
Hence it is better to put all bottles at once and get the fridge
warmer to have a maximum heat pumpimg capacity from the heat pump.
Heat sharing rate within the fridge may also be an important issue, but there are no data available to measure how important. If it is really low, thus leaving an important temperature gradient in the fridge, it may be useful to exchange the position of bottles, so as to have the warmer part of the load
near the heat pump and have work with highest possible coefficient of performance.
Precise figures about the load do not matter very much. However, a load with large heat capacity will take longer to cool and will thus allow more time for heat sharing.
In the second part below, we prove formally that all bottles should be cooled at once, and we use the understanding to discuss the heat sharing issue in some more depth. The variability of the coefficient of performance with temperature is central to this analysis.

FORMAL STATEMENT AND PROOF
A refrigerator is a Carnot machine functionning as a heat exchanger,
where we are interested in removing heat from the low-temperature
reservoir, using work from an engine that provides compression.
The effectiveness, or coefficient of performance, noted here $Z$, is
defined as $Z=Q_c/W$ where $W$ is the work provided and $Q_c$ is the
amount of heat extracted from the cold source (the refrigerator) with
that work. If we note $Q_h$ the amount of heat delivered to the hot
source (outside the refrigerator), we have the equality $Q_h=W+Q_c$.
For an ideal Carnot cycle, we have
$Z_{ideal}=Q_c/W=Q_c/(Q_h-Q_c)=T_c/(T_h-T_c)$ where $T_h$ and $T_c$ are the
temperatures of the hot and cold source. (see http://en.wikipedia.org/wiki/Coefficient_of_performance).
Of course, the actual coefficient of performance $Z$ is less that the
Carnot ideal. Short of knowing its specific, we will only assume that,
like the ideal coefficient, it depends monotonically on the
temperature $T_c$ of the cold source, the hot source (outside the
refrigerator) being considered at constant temperature. Hence we only
assume that the coefficient of performance $Z$ is a strictly
increasing function of (cold source) temperature, i.e., such that
$T_1< T_2\ \Rightarrow\ Z(T_1) < Z(T_2)$
We also assume that the mechanical power available for compression is
invariant, i.e. does not depend on the temperature of the sources, at
least within the range of temperatures considered.
Finally, we also assume that the heat capacity of the refrigerator
itself is negligible, and that heat sharing within the refrigerator
takes negligible time compared to cooling time so that the
content may be considered to have uniform temperature. These later two
assumptions will be discussed afterwards.
With the above assumptions, given two masses $m_1$ and $m_2$ to be
cooled in the cold source, it is faster to cool booth simultaneously
than to try to cool one first and later add the second one. It also
consumes less energy.

PROOF
Cooling a mass $m$
Actually, the formulae above are about heat and work increments. This
is necessary since $Z$ is temperature dependant, and
temperature may vary. Also, since we intend to analyze the system from
the point of view of the cold source, the heat increments are actually
removed from that source and must be countd as negative.
So we can write $Z= -dQ_c/dW$, or  $dQ_c/dW=-Z$.
The power provided by the compressor is a constant $P=dW/dt$.
Hence $dQ_c/dt= (dQ_c/dW)\times(dW/dt)=-Z\times P$.  
On the other hand we know that removing heat reduces the temperature
according to the formula $\Delta Q=-cm \Delta T$, where $m$ is the mass
being colled and $c$ is the specific heat for the substance
constituting that mass.
Hence we have $dQ_c/dt= cm(dT_c/dt)$.  
Combining the two formulae, we get $dT_c/dt=-ZP/cm$.
But we cannot resolve this equation since $Z$ is an unknown function of
$T_c$.
What we know is that $Z(T_c)$, $P$, $c$ and $m$ are strictly positive
values. So $dT_c/dt$ is strictly negative. Hence $T_c$ will
decrease with time. Since the function $Z(T_c)$ is a strictly
increasing function, its value will also decrease with time, and hence
the absolute value of the derivative $dT_c/dt$ will also decrease with
time.
Hence the graphical representation of the evolution of the temperature
will look like the red curve in the figure, where $T_0$ is the initial
temperature at time $t_0$.
Cooling the same mass $m$ in two steps
If we consider cooling independently (in an identical refrigerator) another mass $m_1$, smaller than
$m$, with initial temperature $T_0$ we get another curve, like the curve in blue in the left part of
the figure up to point C, corresponding to the equation
$dT_c/dt=-ZP/cm_1$. It is below the red curve because the smaller mass
$m_1$ cools faster than $m$. Formally, if we draw a horizontal line
like the line cutting both curves in A and B, this corresponds to the
same temperature for both curves, hence to a common value of the
coefficient of performance $Z$. Then $m_1<
m\ \Rightarrow\ (dT_c/dt)_A<(dT_c/dt)_B$. Since this is true for any
value of the temperature $T_c$, it confirms that the blue curve for
$m_1$ decreases faster than the red curve for $m$.
Suppose now that at time $t_2$ the mass $m_1$ has been cooled to
temperature $T_2$ corresponding to point C below the red curve.
We add to $m_1$ another mass $m_2$ such that
$m=m_1+m_2$, the mass $m_2$ being at the initial temperature $T_0$.
The mass $m_2$ being warmer than $m_1$ will share its heat with $m_1$
(in negligible time according to our hypothesis) so that both reach
the temperature $T_1$ corresponding to point D, and pursue cooling.
At any time between $t_0$ and $t_2$, the temperature of $m_1$ (blue)
is less than the temperature of $m$ (red). Hence the refrigerator
works with a lower coefficient of performance $Z$ for $m_1$ than for
$m$, and less heat has been removed from the refrigerator containing
$m_1$ than from the refrigerator containing
$m$ at time $t_2$. When we introduce the mass $m_2$ with $m_1$, the
total heat introduced in the refrigerator is that of $m_1+m_2$ at
temperature $T_0$. This is exactly the same as the heat introduced in
the refrigerator containing $m$. Since less heat was removed from the
$m_1+m_2$ refrigerator at time $t_2$, it is at a higher temperature
that the $m$ refrigerator. Hence the point D is above the red curve.
The $m_1+m_2$ refrigerator now contains the same mass as the $m$
refrigerator. Hence it will follow an identical curve. But it is at
temperature $T_1$ that was attained earlier, at time $t_1$ by the $m$
refrigerator. So the right part of the blue cooling curve for $m_1+m_2$,
starting at point D is the same as the right part of the red coling curve
for $m$ starting at point B, translated by a duration $\Delta
t=t_2-t_1$.
Conclusion
The masses $m_1+m_2$ will always reach any temperature with a delay
$\Delta t$ after the mass $m$ has reached it.
Actual figures would be required to be more precise.
Given the problem, the cooling engine will be working at maximum power
to get the fastest possible cooling in both cases. Then it is obvious
that the faster solution is also the most economical energetically.
This assumes either that the cooling is started just at the right
time, or that the cooling power is reduced once the right temperature
has been attained.
These results are based exclusively on our assumptions, independently
of any actual figures.
We will now discuss some of these assumptions.
DISCUSSION
Heat capacity of the refrigerator
We have assumed that the heat capacity of the refrigerator itself is
negligible. We should however analyse its effect. We first note that
the objective is to extract heat from a given number of bootles to
bring them from $T_0=30^{\circ}C$ to $T_f=6^{\circ}C$. That
corresponds to a precise amount $Q$ of heat to be removed,
independently of the process used for that purpose. 
If the refrigerator itself is initially at temperature $T_f$, it will at
the start share heat with the mass to be cooled, thus warming up and
cooling the bootles. But
then it will have to be cooled down to $T_f$ again, so that it net
contribution to the cooling process will be null, and the same amount $Q$
of heat has to be removed by the heat pump. However, by sharing heat
at the beginning, it induces an early cooling, thus making the whole
heat removal process operate at a lower temperature, hence with a
lower coefficient of performance $Z$.
The net effect of the heat capacity of the refrigerator is thus to
provide some early cooling, which can sometimes be considered an
advantage, but at the expense of a lower effective coefficient of
performance $Z$. These effects increase with the heat capacity of the
refrigerator.
Note that if the initial temperature of the refrigerator is below the
targeted final temperature $ T_f$, the difference multiplied by the
heat capacity is a net contribution to the refrigeration process,
though the loss on the coefficient of performance remains.
Hence it is better not to have anything else in the refrigerator, even already cooled, unless it is cooled to a much lower temperature than the final temperature $ T_f$ intended for the bottles.
Heat sharing rate
As we have seen from the previous discussion, the main objective is to
remove a given amount $Q$ of heat, and the effectiveness of the
removal decreases as temperature is lowered. If the rate of heat
sharing within the refrigerator is small, the volume near the cooling
system will cool faster, thus reducing the coefficient of performance,
i.e. the rate of heat removal.
Hence, ensuring the best possible heat sharing can help in all
circumstances. It should be noted that direct sharing between
cylindrical bottles will be reduced to a minimum: just one line of
contact.  So, if space allows, it is probably preferable to help air
circulate between the bottles. Keeping the bottles in vertical
position will help if the refrigerator has grid shelves that let air
through, rather than glass shelves. And, of course, the bottles must
be unpacked.
Opening the door to quickly exchange bottles so that the warmer ones
are placed near the cooling system will improve the coefficient of
performance and reduce cooling time. It may have some
cost in warming the refrigerator, but that is less important if
the heat capacity of the load to be cooled is large (actual
measurements would be useful).
Exchanging bootles will also avoid having to cool those close to the
cooling system below the required temperature $ T_f$ in order to have
all the bottles temperature at least as low as $ T_f$.
A: It depends on the volume of your fridge versus the amount of bottles you have. Either way it is always best to keep the bottles that are being cooled as far apart as possible in order to maximize surface area to release heat. 
If your fridge can barely fit all the bottles at once then it would be better to put a few bottles at a time in the fridge, wait for them to cool close to 6 ∘C, pack them tightly into a corner, and then put more bottles to cool. If your fridge can easily fit all the bottles then you could just put all the bottles in at once so that they are as far apart as possible.
The reason the size of the fridge matters is because if the bottles are packed tightly then the outer bottles would insulate the inner bottles from the cold fridge air.
A: Given limited time I believe the best strategy would be to get a tub of ice water and put as many beers as you can in there. Those should cool sufficiently between 10-20 min. Put the rest of the beer in the fridge while it's chilling. After the ice beer has chilled, put it in the fridge. The large mass of cold beer will keep the temperature from dropping as much when you open the door. Trade the chilled beer for any other beer that hasn't been cooled yet.
If you are really in a rush you can borrow a technique used when making ice cream. Add salt to the ice water and it will drop the temperature of the ice water from 0C to ~-21C. Beer will chill in 2 minutes at that temperature. Just be careful because it's really cold.
EDIT:
I should also include a strategy for using only the fridge. You want the surface of the beer bottle to be in contact with as cold of a medium as possible. This means you want to avoid a close packed configuration because otherwise the warm beer would be touching other warm beer. Try to arrange the beer with gaps so cold air can circulate. If your fridge is anything like mine, the cold element is at the top of the fridge. This means the first beers to get cold will be at the top. However be absolutely sure to keep the top as loosely packed as allowed because otherwise the cold air won't flow down to the other bottle. If you do arrange bottles in a close packed configuration, make sure the bottle are vertical and not horizontal. Horizontal bottles will leave no gaps for the air to fall whereas vertical bottles will have gaps for the air to fall through to the other bottles.
A: The key to answering this question is understanding whether it is your refrigerator's maximum heat exchange rate or the beer's cooling rate that is limiting.  Your beer needs to lose about ~100J/g of heat.  Your refrigerator is perhaps 15 $\mathrm{ft}^3$, of which maybe a third is actually beer, so you'll have about $5\cdot\left(12\cdot2.54\right)^3 = 142 \mathrm{kg}$ of beer in there requiring 14 MJ of heat transfer.  At 200W and a typical coefficient of performance of about 5 (I think this is rather optimistic but nobody actually reports this number as far as I can tell!), that's still about 14,000 seconds to cool off the entire load, or about four hours.  Recommendations for chilling wine usually are in the 1-2 hour range, and beer bottles are a lot smaller.
So this is likely going to be dominated by your refrigerator's maximum heat transfer rate.  You can put the beer pretty much anywhere and it will turn out the same.  If it turns out that there's a really strong temperature gradient across the refrigerator, you might want to cycle the bottles between the coldest and warmest spots.
A: The fridge loses cold air every time you open it's door. So the best strategy involves minimizing the number of times you open it's door. This suggests the best strategy is to fill the fridge with beer and leave it shut until the party starts.
Anything else in the fridge will probably be ruined because the 30C beer may raise the internal temperature of the fridge well above the safe level for most perishable products.
A: There are already very good answers, but I will add my two cents here.
There are many variables that play a role in this problem but the crucial one is the heat transfer rate. Increasing the rate of heat transfer by increasing the temperature difference, like using a freezer or a mix of salt and ice does indeed the job faster, but increases the risk of breaking bottles. In your case the most practical way is that you increase the time by filling the fridge, set it to a low temperature (but above zero) and keep it closed for a few hours. 
The solution posted above, using the bathtub, is straight on the spot since you change the thermal conductivity of the medium in which the bottles are by one order of magnitude. I have direct experience with this. I grew up in a place with warm summers. Every now and then during the summer we would host a meeting. We had only one normal-sized refrigerator, so using it was out of the question. We would then place a large (~ 0.5 m$^{3}$) plastic box in the shadow of a tree in the garden during the mid-afternoon. We then placed the bottles and cans in a not so tight packing and filled the rest of the space with some water and ice cubes, that we would buy in two bags of about 15 kg each. The box was then covered with two or three old bed covers. By early night all drinks would be chilly. No advanced tech nor high heat transfer rates but worked beautifully!
