The ninety-sixth question: Given $S=\{(\mathrm{i}, \mathrm{j}) \mid \mathrm{i}, \mathrm{j}=1, 2, \ldots, 100\}$ as the set of $100 \times 100$ integer points on the Cartesian plane. Each point in $S$ is colored with one of four given colors. Find the maximum possible number of rectangles with vertices of four different colors and sides parallel to the coordinate axes.
The 96th problem:
The answer: The maximum value sought is 9375000.
Let the four colors be $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}$. Define
$$
R_{i}=\{(x, i) \mid x=1, 2, \ldots, 100\}
$$

as the $i$-th row of the integer point square.
$$
T_{j}=\{(j, y) \mid y=1, 2, \ldots, 100\}
$$

as the $j$-th column of the integer point square.
A rectangle with four vertices of different colors in $S$ and sides parallel to the coordinate axes is called a "four-color rectangle".
We first prove a lemma: If $\mathrm{R}_{\mathrm{i}}$ and $\mathrm{R}_{\mathrm{j}}$ are exactly the same color at $\mathrm{k}$ positions, then the number of four-color rectangles generated by these two rows does not exceed $\frac{(100-\mathrm{k})^{2}}{4}$.

Proof of the lemma: If $R_{i}$ and $R_{i}$ are the same color at some position, then these two points cannot belong to a four-color rectangle. If $(x, i)$ and $(x, j)$ are colors $A$ and $B$, then only when $(y, i)$ and $(y, j)$ are colors $C$ and $D$ will these four points form a four-color rectangle. Similarly, other cases can be derived.

In 100 pairs of grid points $(x, i), (x, j) (1 \leq x \leq 100)$, let there be $u_{1}$ pairs of grid points with colors $A$ and $B$, $u_{2}$ pairs with colors $A$ and $C$, $u_{3}$ pairs with colors $A$ and $D$, $u_{4}$ pairs with colors $B$ and $C$, $u_{5}$ pairs with colors $B$ and $D$, and $u_{6}$ pairs with colors $C$ and $D$. Then
$u_{1}+u_{2}+u_{3}+u_{4}+u_{5}+u_{6}=100-k$
Thus, the number of four-color rectangles generated by $R_{i}$ and $R_{j}$ is:
$u_{1} u_{6}+u_{2} u_{5}+u_{3} u_{4} \leq \frac{\left(u_{1}+u_{6}\right)^{2}}{4}+\frac{\left(u_{2}+u_{5}\right)^{2}}{4}+\frac{\left(u_{3}+u_{4}\right)^{2}}{4} \leq \frac{\left(\sum_{i=1}^{6} u_{1}\right)^{2}}{4}=\frac{(100-k)^{2}}{4}$
The lemma is proved!
Now, let's return to the original problem.
Let $R_{i}$ and $R_{j}$ be the same color at $k_{i j}$ positions, then $\Sigma_{1 \leq i<j \leq 100} k_{i j}$ is exactly the number of pairs of same-color points with the same x-coordinate. Counting it from the first column, let $T_{j}$ have $a_{j}, b_{j}, c_{j}, d_{j}$ points of colors $A, B, C, D$ respectively, then $a_{j}+b_{j}+c_{j}+d_{j}=100$, and the number of same-color point pairs in $T_{1}$ is: $C_{a_{j}}^{2}+C_{b_{j}}^{2}+C_{c_{j}}^{2}+C_{d_{j}}^{2}$. Thus, we know:
$$
\begin{array}{l}
\sum_{1 \leq i<i \leq 100} k_{i j}=\sum_{j=1}^{100}\left(C_{a_{j}}^{2}+C_{b_{j}}^{2}+C_{c_{j}}^{2}+C_{d_{j}}^{2}\right) \\
\geq \sum_{i=1}^{100} 4 \cdot C_{a+b_{j}+c_{j}+d_{j}}^{4}=400 \cdot C_{25}^{2}=120000
\end{array}
$$

Combining the lemma and (1), the number of four-color rectangles does not exceed:
$\sum_{1 \leq \mathrm{i}<\mathrm{j} \leq 100} \frac{\left(100-\mathrm{k}_{\mathrm{ij}}\right)^{2}}{4} \leq \sum_{1 \leq \mathrm{i}<\mathrm{j} \leq 100} \frac{100\left(100-\mathrm{k}_{\mathrm{ij}}\right)}{4}$
$=2500 \cdot C_{100}^{2}-\frac{\Sigma_{1 \leq 1<i \leq 100} k_{i j}}{4} \leq 2500 \cdot C_{100}^{2}-\frac{120000}{4}=9375000$
Now, construct a coloring method such that the graph has exactly 9375000 four-color rectangles.
Divide the $100 \times 100$ integer points into 16 regions, each region being $25 \times 25$ integer points, and color each region as shown in the figure below.

By the above coloring method, when the first grid point of two rows belongs to the same region, these two rows do not generate a four-color rectangle; when the first grid point of two rows does not belong to the same region, these two rows have 4 pairs of regions forming the same color rectangle, which can generate $4 \times 25^{2}=2500$ four-color rectangles. Therefore, the total number of four-color rectangles in the graph is $2500 \left(25 \times 25 \times C_{4}^{2}\right)=9375000$.
In conclusion, the maximum value sought is 9375000.