A puzzle from the 1997 Ukrainian Mathematical Olympiad:
Cells of some rectangular board are coloured as chessboard cells. In each cell an integer is written. It is known that the sum of the numbers in each row is even and the sum of numbers in each column is even. Prove that the sum of all numbers in the black cells is even.

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To begin with, notice that the sum of the numbers in the white cells in any given row or column must have the same parity (odd or even) as the sum of the numbers in the black cells. That’s necessary because we know that the sum of the whole row (or column) is even: If the white total is even, then the black total must be even, and if the white total is odd then the black total must be odd.
Since that’s so, and since we’re interested only in parity, then in evaluating the black squares over the board as a whole we can make that substitution — in a given row or column, we can substitute the parity of the white sum for the parity of the black sum.
Then let’s calculate the sum of the board’s black cells by considering each column in turn, and we’ll choose to sum the black cells in the oddnumbered columns and the white cells in the evennumbered columns. But that’s just the same as summing every second row on the board, and we already know that each of those totals is even. So the final sum must be even.
(Via R.E. Woodrow, “Olympiad Corner,” Crux Mathematicorum 29:2 [March 2003], 87103.)
