Question 2 If it is only required that no 5 stones appear consecutively in horizontal or vertical lines; then, on an $m \times n$ $(m \geqslant 5, n \geqslant 5)$ board, what is the minimum number of stones that need to be removed to possibly meet the requirement?
Reviewing the previous scenarios (1) and (2), we find that their discussion process is still valid for question 2. For scenario (3), on a $7 \times 8$ chessboard, we can only prove: at least 10 chess pieces need to be removed. Therefore, on an $m \times n$ chessboard, the minimum number of chess pieces that need to be removed becomes:
$$
\left\{\begin{array}{ll}
{\left[\frac{m n}{5}\right]-1,} & \text { if } m \text { and } n \text {, when divided by } 5, \\
{\left[\frac{m n}{5}\right],} & \text { otherwise. }
\end{array}\right.
$$

Below, we construct a method that satisfies the above requirements. Here, we use a "diagonal" method: remove all chess pieces whose coordinates satisfy $51(i, j)$, see Figure 6.
On the chessboard in Figure 6,

by appropriately cropping an $m \times n$ sub-chessboard, the requirement can be met.