4. In a Cartesian coordinate system, draw all rectangles that simultaneously satisfy the following conditions:
(1) The sides of these rectangles are parallel or coincide with the coordinate axes;
(2) All vertices of these rectangles (repeated vertices are counted only once) are exactly 100 integer points (points with both coordinates as integers are called integer points).

Question: What is the maximum number of such rectangles that can be drawn? Explain your reasoning.
4. First, prove that the number of such rectangles does not exceed 2025.

Take any 100 integer points. Let $O$ be one of the 100 integer points we have chosen. We call a rectangle "good" if $O$ is one of its vertices, the other three vertices are also chosen from these 100 integer points, and the sides are parallel or coincide with the coordinate axes.
We will prove: There are at most 81 good rectangles.
In fact, draw lines $l_{1}$ and $l_{2}$ through $O$ parallel to the coordinate axes, and let $l_{1} \backslash\{O\}$ contain $m$ points chosen from the 100 integer points, and $l_{2} \backslash\{O\}$ contain $n$ points chosen from the 100 integer points. Let $P$ be one of the 100 integer points chosen, and not on $l_{1}$ and $l_{2}$. Then, at most one good rectangle can have $P$ as one of its vertices. There are at most $99-m-n$ such points, and each good rectangle must have one vertex as such a point. Therefore,
(i) If $m+n \geqslant 18$, then the number of good rectangles is at most $99-m-n \leqslant 81$;
(ii) If $m+n \leqslant 18$, consider the point pairs $(R, Q)$, where $R \in l_{1} \backslash\{O\}$ and $Q \in l_{2} \backslash\{O\}$. Each pair $(R, Q)$ can form at most one good rectangle, so the number of good rectangles is less than or equal to
$$
m n \leqslant m(18-m) \leqslant 9 \times 9=81 \text { . }
$$

In summary, for any point $O$ among the 100 integer points chosen, the number of good rectangles with $O$ as one of its vertices is at most 81. Therefore, the number of rectangles that meet the conditions is less than or equal to $\frac{81 \times 100}{4}=2025$ (here we divide by 4 because each rectangle has 4 vertices).
Below is an example of 2025 such rectangles.
Consider the point set
$$
A=\{(x, y) \mid 1 \leqslant x \leqslant 10,1 \leqslant y \leqslant 10, x y \in \mathbf{N}\} .
$$

By choosing 100 points from the point set $A$, we can exactly draw 2025 rectangles that meet the problem's conditions.
Therefore, the maximum number of such rectangles that can be drawn is 2025.