Let the set $X=\{1,2, \cdots, 100\}$. Ask: Does there exist 1111 different subsets of $X$, such that the intersection of any two of them (allowing the same subset to be chosen twice) has a number of elements that is a perfect square?
Note: A perfect square refers to $0,1,4,9,16, \cdots$.
Let $X$ be equivalent to
$$
\{1,2, \cdots, 10\} \times\{1,2, \cdots, 10\} \text {, }
$$

then the 1024 subsets given by $\{S \times S \mid S \subset\{1,2, \cdots, 10\}\}$ satisfy the requirement. This is because
$$
(S \times S) \cap(T \times T)=(S \cap T) \times(S \cap T),
$$

Additionally, the remaining 90 single-element sets can still be added.
Therefore, 1114 is achievable, and selecting 1111 of them is sufficient.