When is it more efficient to blow air over a wet laundry in order to dry it - when it's wetter or when it's drier? If I want to speed the drying of laundry and allocate for it one hour of fanning - should I use it just after I hang it to dry or several hours later?
When we want to cool hot tea it's better to add the cold milk just before drinking it. Is it the same also in this case? 
Edit:
I like your answers, all in favor of the "in the beginning" option. Is it always the better option in cases where doing nothing is also a solution, i.e. when we don't really have to invest energy in order to get the desired outcome (unlike the coffee case where the milk is usually colder than the air)?
 A: I haven't done the calculations on this, but I think viewing this from a standpoint of energies would get you a long way.
Your example with the tea and the milk is true because the milk absorbs some of the energy of the hot tea. Essentially:

Tea is hot relative to surroundings and milk
Larger difference in temperature = larger flow of heat = faster cooling down
Adding milk at the start absorbs energy (temperature) from hot tea, slows the cooling process
Adding milk at the end absorbs less energy, but cooling process has run at maximum speed until then
Net result is faster cooling down when milk is added at the end

The blowing fan with the wet laundry works a bit differently.

Laundry is humid compared to surroundings
Larger difference in humidity = higher rate of evaporation = faster drying
Turning on the fan at the start blows away humid air near the laundry, increasing the humid-dry gradient near the laundry
Drying process speeds up
Turning on the fan at the end blows away humid air near the laundry (but less than when it was more wet), increasing the humid-dry gradient but less so than when it was more wet
Drying process speeds up less
Net result is faster drying when fan is turned on at the start

I am assuming the fan blows air at room temperature. If it blows hot air the result is the same, but you won't get that cuddly soft and warm feeling laundry at the end. So if you want that, turn on the hot air near the end of the drying process and cozy up to your warm dry laundry with a cup of hot chocolate.
A: Okay, so the wetness ($w$) starts at 1, then every second, let's say it multiplies by $1-d$, with d being an infinitesimally small number. So we get a nice, smooth half-life curve (more generally known as exponential decay). Then, we can see that, without even having to go into anything resembling exact numbers, if the fan makes this accelerate to the stable-state, then you should use it immediately. If you imagine the ball-rolling metaphor (where it starts off with lots of velocity but gradually slows down as it approaches the equilibrium), this is going to pause reality then push it further without affecting it's velocity, which is more useful at the beginning than at the end.
Edit: I've thought long and hard over this, and if you just enter $y=\frac12^x$ into a graphical calculator (like Desmos), you can see that it won't do what I said at all. It will just accelerate the process. Say it pushes air particles so that twice as many interact with (and steal energy from/dry) the tea/laundry, then it would simply make it move faster along this graph. If you look at it through logarithmic eyes (with $y=log_2(\frac12^x)$), you'll see that it's cancelling itself out, showing that the orders of magnitude of difference decrease linearly, which means that the placement of fan-blowing for $n$ time would have no difference in the end. But this is in an idealised world, I'm not a laundry-cleaning expert, and the best method is to try getting data for yourself, with a thermometer, instead of asking a bunch of bored physicians, mathematicians and programmers.
TL;DR: Do it at the start. Edit: THIS IS ALL MEANINGLESS.
A: This case is quite similar to the heating case in some ways, but not others.
Thermal gradients behave quite similarly to evaporation concentrations.  That is, the rate of evaporation is greater if the difference in concentration is greater.  The rate is also increased by convection, the same as with heating.
A big difference is what we are doing in this situation compared to adding milk to a warm cup.  In that case, you have a low temperature mass you can add, and you are just looking for the right time to add it.  For that to be equivalent here, we would need some sort of absorbent object to put into the mix, which is obviously not the case here.
The analogous heat transfer question to ask here is "If I can only blow on the coffee once, should I blow on it right after I pour it, or right before I drink it.
Given that we are blowing for a set amount of time, and the greatest mass transfer occurs when the gradient is greatest and there is convection, to maximize the mass transfer over the time being blown, I would choose to blow on it when it is the wettest.  This at very least will get you closer to dry far earlier.
A: You ask at what time is the fan most efficient, which makes this problem different posed than the cup of tea and milk. We can see intuitively that if you wait a very long time, the fan will do almost nothing, because the laundry is almost dry by then anyway. You get a smaller change in dryness the longer you wait, so it is less efficient the later you run the fan. 
You can actually see that's the case for the tea and milk, too - if you wait a long time before adding the milk, you might have a lower final temperature, but the tea will be mostly cool by then, so the milk has done relatively little to cool the tea. If you had added it at the start, the milk would have cooled the tea significantly all by itself. So, the milk-adding is actually more efficient the earlier you do it, if we measure efficiency as "degrees cooled per oz".
