Place the chessboard so that the top left-hand square is black. The $64$ squares have $8$ columns that can be labelled $1, 2, 3, …, 8$ from left to right and $8$ rows that can be labelled $1, 2, 3, …, 8$ from bottom to top in the usual manner. The white squares in the odd-numbered rows are in different columns to those in the even-numbered rows, so we label each of the former with the number $1$ and each of the latter with the number $2$. Similarly, we label each of the black squares in the odd-numbered rows with the number $3$ and the remaining black squares with the number $4$. There are $4$ odd-numbered rows, each containing only the numbers $1$ and $3$, so if we let $n(1)$ be the number of pieces on a square labelled $1$ and $n(3)$ the number of pieces on a square labelled $3$, then we must have $n(1) + n(3) = 4$. Similarly, from the odd-numbered columns, we get $n(1) + n(4) = 4$. This gives $n(3) = n(4)$, meaning that $n(3) + n(4)$ must be even, that is, these has to be an even number of pieces on the black squares.