A rectangle has all its side lengths odd integers. We partition it into several small rectangles, such that each small rectangle has all its side lengths positive integers. Prove that among these, there is at least one small rectangle whose distances to the four sides of all have the same parity.
(Here, the distance from a small rectangle to a side of is defined as the perpendicular distance between and the side of the small rectangle parallel to that is closer to .)
Let the length and width of be respectively. Divide the rectangle into a grid of unit squares, and color them alternately black and white in a chessboard pattern. Note that since are both odd, the four corner squares must be colored the same color; without loss of generality, assume it is black.
For each rectangle, we call it black (white) if and only if all four of its corner squares are black (white); otherwise we call it gray. Note the following facts:
(i) A gray rectangle contains the same number of black and white squares;
(ii) A black rectangle contains one more black square than white squares;
(iii) A white rectangle contains one more white square than black squares.
Now, is black, so it contains one more black square than white squares; combined with the above facts, we know that there must be at least one small rectangle that is black. Now let the distances from this rectangle to the four sides, starting from the top and proceeding clockwise, be respectively. Since the small rectangle and the top-right corner of are both black, it is easy to see that and must have the same parity. In the same way we can show that and , and , and all have the same parity, so this black rectangle's distances to the four sides all have the same parity. This completes the proof.