Solution:

Color the chessboard with the usual chessboard coloring. Note that bishops can only attack knights on the same colored squares, while knights can only attack bishops on different colored squares. Let $B_{B}, W_{B}, B_{N}, W_{N}$ denote the number of bishops on black colored squares, bishops on white colored squares, knights on black colored squares and knights on white colored squares respectively.

Since removing a knight on a white colored square will cause a leave a bishop on a white colored square with no knight to attack, $W_{N} \leq W_{B}$. Similarly, $B_{N} \leq B_{B}, B_{B} \leq W_{N}, W_{B} \leq B_{N}$. Combining the inequalities, we find that $W_{N}=W_{B}=B_{N}=B_{B}$. Hence $4 \mid n$.

To construct $n=4$, we can place bishops at $(1,0),(2,2)$ and knights at $(0,1),(3,1)$. We can easily extend this to any multiple of $4$ by placing this configuration as many times as required across the grid, making sure that we avoid the (finitely many) diagonals that existing bishops occupy.