You have 64 chess queens which come in 8 colors, with 8 queens per color. The goal is fill a chessboard with these queens so that any two queens of the same color cannot attack each other, even when allowed to move through differently colored queens. This means no two similarly colored queens may share a rank, file or diagonal.
In other words, can you fill a chessboard with 8 disjoint solutions to the classic Eight Queens puzzle?
What about with $n^2$ queens on an $n\times n$ board?
It is not possible on an 8x8 board.
The eight queens of each colour have to form a solution to the standard eight queens problem. There are only 12 such patterns (plus their rotations/reflections). We need both main diagonals to contain one queen of each colour, but only six of the 8-queen patterns have queens on both main diagonals. Unfortunately all six of those patterns also use up a square diagonally adjacent to a corner. There are only four such squares, but we would need to combine eight of those patterns to form a full solution.
On my puzzle website there is an old puzzle called Hoo-Doo which is essentially this same puzzle.
Some other board sizes:
A 7x7 solution:
1 2 3 4 5 6 7
6 7 1 2 3 4 5
4 5 6 7 1 2 3
2 3 4 5 6 7 1
7 1 2 3 4 5 6
5 6 7 1 2 3 4
3 4 5 6 7 1 2
A 5x5 solution:
1 2 3 4 5
4 5 1 2 3
2 3 4 5 1
5 1 2 3 4
3 4 5 1 2
There is no solution for smaller boards or for 6x6.
If $n$ is coprime to 6, then you can make a solution for the $n\times n$ board similar to the two solutions above. Each row is the row above cyclically shifted right two steps. I suspect the other board sizes have no solution, but have not been able to prove that.
Here is a a demonstration.
It is not possible
Take the following board:
It is a solution for castles, but not queens. (If queens couldn't move diagonally, this would be an acceptable solution)
So to fix the diagonal problems we can switch some of the queens. But if you switch enough queens you will result in a horizontal or vertical error, hence it is impossible
