No have been solved this yet

We show this is impossible for even $2n$. Label the rows of the grid $1, 2, \dots, 2n$ and take colorings of the square in column $x$ and row $y$ based off $x + y \pmod{2n}$, which has $2n$ elements for each residue. Then we claim that for any nonthreatening rook config, two have the same colored square. This is equivalent to showing that if we take permutations $a_1, b_1$ of $1, 2, \dots, 2n$, then two of $c_i = a_i + b_i$ are the same $\pmod{n}$. FTSOC suppose not. Then \[ \sum_{i=1}^{2n} i \equiv \sum_{i=1}^{2n} c_i = \sum_{i=1}^{2n} (a_i + b_i) = 2\sum_{i=1}^{2n} i \pmod{2n} \]which is a contradiction.