Maximum time difference between clocks in a gravity field From Surely you must be joking, Mr Feynman.

You blast off in a rocket which has a
  clock on board, and there's a clock on
  the ground. The idea is that you have
  to be back when the clock on the
  ground says one hour has passed. Now
  you want it so that when you come
  back, your clock is as far ahead as
  possible. According to Einstein, if
  you go very high, your clock will go
  faster, because the higher something
  is in a gravitational field, the
  faster its clock goes. But if you try
  to go too high, since you've only got
  an hour, you have to go so fast to get
  there that the speed slows your clock
  down. So you can't go too high. The
  question is, exactly what program of
  speed and height should you make so
  that you get the maximum time on your
  clock?
This assistant of Einstein worked on
  it for quite a bit before he realized
  that the answer is the real motion of
  matter. If you shoot something up in a
  normal way, so that the time it takes
  the shell to go up and come down is an
  hour, that's the correct motion. It's
  the fundamental principle of
  Einstein's gravity--that is, what's
  called the "proper time" is at a
  maximum for the actual curve.

How is this result proven?
 A: As Raskolnikov says, it's the principle of least action (more precisely extremal action but let's put that off till later). Consider the task of finding the shortest path between two points in Euclidean space. We call such a special path geodesic and it's obvious that here it is just plain old line segment between the two points. But to prove this statement is not so trivial (one needs calculus of variations at hand), so let's concentrate on just the intuitive picture.
If you try to deform the line a little to make the total path shorter, you'll find it's not possible. Reason is that infinitesimally, each such deformation will correspond to a triangle with longest side $c$ being the part of the line and $a$ and $b$ lines comprising the perturbed path. As per basic Euclidean geometry we have $c < a + b$.
In Minkowski space-time, the picture is very similar. Straight lines are again geodesics although now they extremize space-time intervals or equivalenty proper time along the path (at least if the particle is massive). Again, if you repeat the above argument you will obtain an inequality but with the sign reversed because time component has different signature than the space components. That means that time will be maximized along the geodesic.
The above can be generalized to curved spaces and space-times. E.g. on a sphere you'll find the shortest path between two points are parts of great circles. And in fact, our argument will work too because infinitesimally sphere looks like a plane (as I am sure you know from your day to day experience living on the surface of Earth) so our arguments holds. And similarly for space-times and proper times.
A note on the word extremal is in order. If you inspect the sketches of the proofs I made, you'll see that we always make only local (infinitesimal) perturbations around the extremal paths to see if there is something shorter/longer around. That means that we only find local extrema. Also, these paths need not exist or be unique between every pair of points (consider the sphere again and all the geodesics between north and south poles). This is good to keep in mind but I won't delve into general mathematical theory as it's messy and not needed usually.
A: To elaborate on Marek's (correct) answer, since it seems that math is the issue that @Casebash is having:
Start with an integral representing the time elapsed on the moving observer's clock in terms of the stationary observer's coordinates (I'm supressing G and c below.  Feel free to replace all $t$'s with $c\,t$'s, and all $M$'s with $\frac{GM}{c^{2}}$'s):
$$\begin{equation}
\Delta \tau = \int d\tau \sqrt{{\dot t}^{2}(1-\frac{2M}{r})- \frac{{\dot r}^{2}}{1-\frac{2M}{r}}- r^{2} {\dot \theta}^{2} - r^{2}\sin^{2}\theta {\dot \phi}^{2}}
\end{equation}$$  
Where $(t,r,\theta,\phi)$ are all considered to be functions of $\tau$ denoting the moving observer's position at time $\tau$.  Our problem amounts to looking for the path beginning at $(t_{0},R,\Theta,\Phi)$ and ending at $(t_{f},R,\Theta,\Phi)$ that maximizes this integral, subject to the constraint that the quantity under the square root will be equal to $1$ when the calculation is complete.  To make our calculations easier, note that the square root is a monotonic function over all its domain, so we might as well maximize
$$\begin{equation}
\frac{1}{2}\int d\tau \left({\dot t}^{2}(1-\frac{2M}{r})- \frac{{\dot r}^{2}}{1-\frac{2M}{r}}- r^{2} {\dot \theta}^{2} - r^{2}\sin^{2}\theta {\dot \phi}^{2}\right)
\end{equation}$$  
which is considerably simpler.  It would take a whole page to carefully vary (meaning that we basically, but not quite, take the derivative of the function with respect to) this with respect to the four independent functions.  So, I'm going to just give you a taste of the task by varying with respect to $\theta$.  The path of maximum "moving clock" time (heretofore called 'proper time') will be the one for which these variations are zero.  The other four functions follow just as easily.  The variation of this integral with respect to $\theta$ gives us
$$\begin{align}
\delta \Delta\tau |{}_{\theta}&=\int d\tau\left(-r^{2}{\dot \theta}\delta {\dot \theta}-r^{2}\sin\theta \cos \theta {\dot \phi}^{2}\delta \theta\right)\\
0&=\int d \tau \left(-\frac{d}{d\tau}\left({\dot \theta} r^{2}\delta \theta\right) + \left({\ddot \theta}r^{2} + 2 r {\dot r}{\dot \theta}-r^{2}\sin\theta \cos \theta {\dot \phi}^{2}\right)\delta \theta\right)\\
&=\left(\dot \theta(t_{0})r(t_{0})^{2}\delta \theta (t_{0})-\dot \theta(t_{f})r(t_{f})^{2}\delta \theta (t_{f})\right)\\
&\quad + \int d\tau\; r^{2}\left({\ddot \theta} + \frac{1}{r}2{\dot r} {\dot \theta}- \sin \theta \cos \theta {\dot \phi}^{2}\right)\delta \theta
\end{align}$$
The first term, that we obtained by integrating the total derivative, vanishes since the variation of $\theta$ is zero at the endpoints $t_{0}$ and $t_{f}$ (our space of states is paths beginning and ending at a fixed value of $\theta$.  If the variation will be zero, then the remaining stuff under the integral must turn out to be zero if it is going to be a minimum for an arbitrary fixed-point variation (if the integrand isn't zero, just imagine making the variation a hundred kajillion only at the point where it is nonzero--clearly this isn't an extremum).  Therefore, the $\theta$ variable of the maximum proper time path must satisfy ${\ddot \theta} + \frac{1}{r}2{\dot r} {\dot \theta}- \sin \theta \cos \theta {\dot \phi}^{2}=0$.  It turns out that this equation is exactly the same equation you get for geodesic motion of a particle, which is the path followed by an observer being acted upon only by gravity.  A simple calculation of the first integral for any other path (and the fact that it will come out higher) will then show you that this, in fact, is the maximum time path to travel.  
A: There's a way to see the answer in the OP logically, no math. In SR, clocks that "feel" acceleration, like the ground clock does, run slower than floating clocks (ones in free fall) do. Therefore the clock that floats during the entire experiment will elapse the most time. Only a clock "shot up in a normal way", to use your words, floats during the entire experiment. Search for the Non-Mathematical Proof of Gravitational Time Dilation for an elaboration. The falling clock in that experiment is one half of a symmetrical experiment involving a clock "shot up in a normal way".
The clock on the ground is analogous to the traveling twin that thrusts its rocket engines in the twin paradox. The clock "shot up in the normal way" is analogous to the stay-at-home twin that floats "stationary" in space, waiting for the traveling twin to return. Any clock shot up, but not in the normal way, will be a bit less like the stay-at-home twin and a bit more like the traveling twin.
BTW, the reason that a floating clock elapses more time than a clock that "feels" acceleration, when they start and end at the same location, is also given by SR: The clocks move relative to each other, completing trips between when they start and end at the same location. The trip travel distance of the clock that "feels" acceleration is shortened by length contraction predicted by SR. (A ruler that freely falls downward and near the clock on the ground shortens as it falls, as measured in that clock's frame, because the ruler's speed is increasing as measured in that frame. The ruler measures a portion of the remaining distance the clock has to travel.) Therefore that clock travels less distance at the same average speed relative to the other clock, elapsing less time than the floating clock does.
