Answer:
$$
4 \mid n
$$

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.
(A harder variant of the problem involves changing the third condition from "If any piece" to "If any proper subset of pieces". The proof remains unchanged but the construction is more tedious as the pieces must be "connected" without accidentally attacking one another. The question can also be phrased in a non-chess way to obscure the motivation behind the coloring.)
2. [7] Some bishops and knights are placed on an infinite chessboard, where each square has side length 1 unit. Suppose that the following conditions hold:
- For each bishop, there exists a knight on the same diagonal as that bishop (there may be another piece between the bishop and the knight).
- For each knight, there exists a bishop that is exactly $\sqrt{5}$ units away from it.
- If any piece is removed from the board, then at least one of the two conditions above is no longer satisfied.
If $n$ is the total number of pieces on the board, find all possible values of $n$.