TITLE: Rooks on a labeled chessboard
QUESTION [7 upvotes]: An $n\times n$ chessboard is constructed such that the coordinate $(i, j)$ is labeled with $i+j \mod n$.
Example for n = 6:

The goal is place $n$ rooks in the chessboard such that none threaten each other and that the same label is never used twice.
The solution is trivial for odd $n$:

The diagonal can always be used as rooks will never threaten each other along a diagonal and all integers $[0, n)$ will be generated as the next entry's label after $(i, j)$ in the diagonal is $i+1+j+1 = i+j+2 \mod n$, and going up to $2\mod n$, for odd $n$, will generate all numbers (staring from $i = j = 0$).
For all even $n$, though, the diagonal is definitely not possible. Is there another solution, though? I do not think so, and manually attempting to generate one on an even board always seems to fail. I'm finding it quite difficult to prove it is impossible, though. 
Help would be greatly appreciated! Thanks!

REPLY [6 votes]: Suppose $n$ non-attacking rooks have been placed on an $n\times n$ chessboard, and let $R=\{(i,j)|(i,j){\textrm{ is occupied}}\}$ Consider $s=\sum_{r\in R}\left((i+j)\mod n\right)$, the sum of the labels of the locations in $R$, taken mod $n$. For the rooks to be non-attacking, there must be exactly one rook in each row and one in each column, so $s \textrm{ mod } n=\left(\sum_{i=0}^{n-1}i+\sum_{j=0}^{n-1}j\right)\!\!\! \mod n = n(n-1)\!\!\! \mod n = 0$. Therefore $s$ divisible by $n$. If $n$ is even, $(0+1+\cdots+n-1)$ is not divisible by $n$, so the labels on the occupied squares cannot be $0, 1, 2, \dots, n-1$.