The 94th question: On a $200 \times 200$ chessboard, some cells contain a red or blue piece, while other cells are empty. If two pieces are in the same row or column, we say one piece can “see” the other. Suppose each piece can see exactly five pieces of the opposite color (it may also see some pieces of the same color). Find the maximum number of pieces on the chessboard.
Question 94:
Solution: Let the total number of pieces on the chessboard be $\mathrm{S}$.
If two pieces of different colors can see each other, draw an edge between them. According to the condition, each piece exactly draws five edges, so there are $\frac{55}{2}$ edges on the chessboard. These edges are either vertical or horizontal. Suppose the number of horizontal edges is not less than the number of vertical edges, then the number of horizontal edges is not less than $\frac{55}{4}$.

Let the $i$-th row ($i=1,2, \ldots, 200$) have $r_{i}$ red pieces and $b_{i}$ blue pieces, then the $i$-th row has exactly $r_{i} b_{i}$ horizontal edges. Therefore, the number of horizontal edges on the chessboard is $\Sigma_{i=1}^{200} r_{i} b_{i}$. Hence,
$$
\begin{array}{l}
\frac{5 S}{4} \leq \sum_{i=1}^{200} r_{i} b_{i} \Rightarrow S \leq \frac{4}{5}\left(\sum_{i=1}^{200} r_{i} b_{i}\right) \\
\text { On the other hand, we know } \\
S=\sum_{i=1}^{200}\left(r_{i}+b_{i}\right)
\end{array}
$$

For any $i \in\{1,2, \ldots, 200\}$, it is clear that $r_{i}+b_{i} \leq 200$. If $r_{i} b_{i}>0$, then there are red pieces and blue pieces in that row, and each red piece can see exactly 5 blue pieces, so $b_{i} \leq 5$, and similarly, $r_{i} \leq 5$, hence $r_{i}+b_{i} \leq 10, r_{i} b_{i} \leq 25$.

Suppose there are exactly $x$ values of $i \in\{1,2, \ldots, 200\}$ such that $r_{i} b_{i}>0$, then the remaining $200-x$ values of $r_{i} b_{i}$ are all 0. Thus, according to (1), we have:
$$
S \leq \frac{4}{5}\left(\sum_{i=1}^{200} r_{i} b_{i}\right) \leq \frac{4}{5} \times 25 x=20 x
$$

On the other hand, according to (2), we have:
$$
S \leq 200(200-x)+10 x=40000-190 x
$$

Combining (3) and (4), we get:
$$
S \leq \min \{20 x, 40000-190 x\}
$$

By calculation, the maximum value of $\min \{20 x, 40000-190 x\}$ is 3800, and this maximum value is achieved when $x=190$. Therefore, $S \leq 3800$.

Below is an example where the number of pieces is 3800:
The intersections of rows 1 to 5 and columns 11 to 200, and the intersections of columns 1 to 5 and rows 11 to 200 are all filled with red pieces; the intersections of rows 6 to 10 and columns 11 to 200, and the intersections of columns 6 to 10 and rows 11 to 200 are all filled with blue pieces. It is easy to verify that each piece can see exactly five pieces of the opposite color. In this case, the total number of pieces is $5 \times 190 \times 4=3800$. In summary, the maximum value sought is 3800.