If $n$ is odd, it is impossible. Use the checkerboard pattern with green corners. Each of the $n$ rooks attacks $n$ green squares and there are $(n^2 + 1)/2$ green squares. Since $2 \cdot (n^2+1)/2 = n^2+1 > n^2,$ some green square will be attacked by at least $3$ rooks. $2$ of those $3$ rooks must be on the same row or column, which means they attack each other. Suppose $n$ is even and the grid has been colored. For $i=1 \to n,$ place a rook in the $i$th row so that it is not attacked and the number of green squares in its column below it is maximized, with ties broken by selecting the leftmost maximizer. Still trying to prove that this algorithm works.