Intuitive/direct proof that a rectangle partitioned into rectangles each with at least one integer side must itself have an integer side A challenge problem asked to show that if rectangle $R$ with axis-parallel sides is partitioned into finitely many subrectangles $R_1,R_2,\ldots,R_n$ (also with axis-parallel sides), such that each $R_i$ has at least one integer side length, then $R$ must also have at least one integer side length. The only proof I've seen, although simple, is quite unintuitive. Namely you look at the integral 
$$ \int_R e^{2 \pi i (x+y)} dx dy $$
and note that the integral is $0$ if and only if $R$ has an integer side length, and then note that by assumption the integral is $0$ over all the $R_i$, hence summing the integral over all $R_i$ gives that the integral is also $0$ over $R$. Can someone give a more direct/intuitive argument that doesn't use voodoo like a complex-valued integral?
Make the plane into a giant checkerboard grid with squares of size $\tfrac{1}{2}\times\tfrac{1}{2}$. In this problem we consider only rectangles with sides parallel to the axes. It is (intuitively) clear that such a rectangle with a side of integer length covers equal amounts of black and white area.

Let $R$ be a rectangle of size $a\times b$ partitioned into finitely many subrectangles, each having a side of integer length. Note that $R$ then covers equal parts black and white, because each of the smaller rectangles does. Without loss of generality we may assume that the lower left corner of $R$ is at the origin.

Divide $R$ into two rectangles of sizes $\lfloor a\rfloor\times b$ and $(a-\lfloor a\rfloor)\times b$, with the first rectangle having its lower left corner at the origin. The first covers equal amounts of black and white. The second can again be subdivided into two rectangles of sizes
$$(a-\lfloor a\rfloor)\times\lfloor b\rfloor\qquad\text{ and }\qquad(a-\lfloor a\rfloor)\times(b-\lfloor b\rfloor),$$
with the former having one side adjacent to an axis. This rectangle again cover equal parts black and white. The latter rectangle must then also cover equal parts black and white.

Note that this smaller rectangle is positioned in the lower left corner of a unit square, covered by precisely two black and two white squares. For it to cover equal parts black and white, one of its sides must have integer length. But for $a-\lfloor a\rfloor$ or $b-\lfloor b\rfloor$ to be an integer, it must be zero. We conclude that either $a$ or $b$ is an integer, and hence that $R$ has a side of integer length.
I can prove it assuming that all side lengths of subrectangles are rationals of the form $q/p$ where $p$ is a fixed prime. In that case, if we scale the rectangle and subrectangles by a factor of $p$ then we get that all side lengths are integers, and all subrectangles have a side length divisible by $p$, and we want to show that the overall rectangle also has a side length divisible by $p$. But this is trivial: every subrectangle has area divisible by $p$, so the total area of the overall rectangle is also divisible by $p$, and hence at least one side of the total rectangle must have length divisible by $p$.