Serbia doesn’t have different problems for different grades in the National Olympiad. This must be some other competition?

Translate to ignore the superfluous details: we can pick at most $4 \cdot 101^2$ subsets of size 4 from $S$ where any two subsets only intersect in at most two elements. Let's generalize: suppose we can pick $m$ such subsets. Double-count. Consider all pairs $(A, B)$ where $|A| = 3, |B| = 4, A \subseteq B$, and $B$ is one of the chosen subsets. On one hand, counting by $B$, there are exactly $4m$ pairs: each $B$ has exactly 4 associated $A$, since there are 4 ways to pick a subset of size 3 from a set of size 4. On the other hand, there are at most $\binom{100}{3}$ pairs; each $A$ is counted at most once, as otherwise two different $B$'s share the same $A$ and thus intersect in 3 elements. Thus $4m \le \binom{100}{3}$. Counting the number gives $m \le 40425$, while $4 \cdot 101^2 = 40804$.