Proofs: the running in the sun conjecture (I made it up - explained below). Is it true and how can it be proven?

I don't know if it's a problem to answer my own question, but I think I've pretty much figured it out just now:

Essentially, we're asking whether the line integral of a constant vector field always sums to zero along any arbitrary closed path.

Here it is - Line integral: http://en.wikipedia.org/wiki/Line_integral#Path_independence

The "path independence" part of the link above is what enables it. We can represent the constant vector field $F$ of the sun's rays as the gradient of a linear scalar field $G$ (changes linearly with $x$, $y$, and $z$). Then, the path integral is equal to $G(x_0) - G(x_0) = 0$ due to path independence of the line integral of a gradient field ... (see link).

If the body is a cylinder, then the normal vector on any portion of the surface is always at a constant angle from the vector facing forward along the running path (because we assume we're a rigid cylinder or any shape) ... So, although the path integral "dots" the vector field with the forward direction along the path, all side directions keep the same angle with the path throughout the entire path, so, their integral is a constant times the same closed path integral, which is zero.

So, we've proved it. In fact, we've proved it for an arbitrary three-dimensional rigid body shape, along any planar path (paths in three dimensions or higher introduce new considerations and are not as straightforward). If the path crosses itself, as someone pointed out in one of the other answers, we can break it down into two simple paths that don't cross themselves and prove the same way we just did that the integral around each of the two simple paths sums to zero - so the integral around the path that crosses itself also sums to zero.

Actually, we set out to ask the question for a constant vector field, but the argument seems to hold for any kind of field that satisfies $F = \nabla G$ ... ($F$ is the gradient of $G$ - where $G$ is a scalar field). So, if $F$ meets this condition, then, around a closed path, the one side of our body will end up getting as much of $F$ as the opposite side of our body will.

However, we should actually note that this ignores one of the implicit, unstated assumptions of the question: what if indeed we take the sun as a spherical source that's a finite distance away and we walk around it in a circle with the center of the circle right below the sun? The rays will then still be representable as a gradient field of a scalar function (since they follow an inverse power law). So, path independence of the line integral will work in that case too because on either side of the body, we're exposed to the same gradient field. The assumption being violated however is that one side of the body will block the rays from reaching the opposite side - i.e. we are an opaque object. Because of this, then, it seems only to work for parallel constant rays ...

To clarify, this works if the path is planar, and the vectors normal to the body are parallel to that plane. For example, sun exposure of the top of the head and one's back will not be the same, but exposure of left shoulder will be the same as that of the right shoulder, and the front will be exposed the same as the back, regardless of where the sun is and regardless of the specific path ... It may or may not work in general with non-planar paths, but if it does, we would need certain conditions on the orientation of the rigid body as it follows the path, and it still won't be true for all surfaces of the body (comments on this point are welcome).

Let the sun be along the $+x$ axis and your starting position be the origin. Let your speed be $1$. We can see that your front axis and rear axis must receive equal amounts of sun by considering the fact that you return to $x=0$. If you move at angle $\theta$ to the $x$ axis for a distance $L$, your front receives $L \cos \theta$ of sun and you net motion in the $+x$ direction is the same. As you return, your rear axis will also get $L \cos \theta.$ Similarly your left shoulder gets sun when you move in $-y$, your right shoulder the same amount when you move in $+y$

For a point an angle $\phi$ from your front toward your right shoulder, it gets an amount of sun proportional to your movement in the $\phi-\frac \pi 2$ direction and the opposite point gets sun proportional to your movement in the $\phi + \frac {3 \pi}2$ direction-again these are the same as you return to the origin.

Crossings are not a problem. Just solve it for one loop at a time.