1. Let the dimensions of the rectangle R be m×n, where both m and n are odd integers.
2. Consider a coordinate system where the bottom-left corner of R is at (0,0), the bottom-right corner is at (m,0), the top-left corner is at (0,n), and the top-right corner is at (m,n).
3. We will use a chessboard coloring argument. Color the rectangle R like a chessboard, where the bottom-left corner (0,0) is colored black.
4. Since m and n are odd, the coloring will ensure that the four corners of R are all black.
5. Now, consider any small rectangle within R with integer side lengths. Let the coordinates of the bottom-left corner of this small rectangle be (x1,y1) and the top-right corner be (x2,y2).
6. The distances from the four sides of R to this small rectangle are:
- Distance to the left side: x1
- Distance to the right side: m−x2
- Distance to the bottom side: y1
- Distance to the top side: n−y2
7. We need to show that at least one of these small rectangles has all four distances either all odd or all even.
8. Observe that the coloring of the chessboard ensures that any small rectangle with integer side lengths will have its corners either all on black squares or all on white squares.
9. If the small rectangle has its corners on black squares, then both x1 and y1 are even, and both x2 and y2 are odd (since m and n are odd).
10. Therefore, the distances:
- x1 (even)
- m−x2 (even, since m is odd and x2 is odd)
- y1 (even)
- n−y2 (even, since n is odd and y2 is odd)
11. If the small rectangle has its corners on white squares, then both x1 and y1 are odd, and both x2 and y2 are even.
12. Therefore, the distances:
- x1 (odd)
- m−x2 (odd, since m is odd and x2 is even)
- y1 (odd)
- n−y2 (odd, since n is odd and y2 is even)
13. In both cases, we have shown that there exists at least one small rectangle within R whose distances from the four sides of R are either all odd or all even.
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