Junior Turkish Mathematical Olympiad

Suppose $S_1$ solved $Q_1, \dots, Q_k$, but not $Q_{k+1}$ where $k > 4$. Since $Q_{k+1}$ is solved by exactly 4 students, and $Q_i$ and $Q_{k+1}$ are solved by exactly one student for each $1 \le i \le k$, there must be another student who solved two of the questions $Q_1, \dots, Q_k$ besides $S_1$, a contradiction. Therefore $k \le 4$. Since $Q_1$ is solved by exactly four students, the number of questions cannot be more than $1+4 \cdot 3 = 13$.

On the other hand, an exam with 13 questions where
$Q_1$ is solved by $S_1, S_2, S_3, S_4$; $Q_2$ is solved by $S_1, S_5, S_6, S_7$; $Q_3$ is solved by $S_1, S_8, S_9, S_{10}$; $Q_4$ is solved by $S_1, S_{11}, S_{12}, S_{13}$; $Q_5$ is solved by $S_2, S_5, S_9, S_{13}$; $Q_6$ is solved by $S_2, S_6, S_{10}, S_{11}$; $Q_7$ is solved by $S_2, S_7, S_8, S_{12}$; $Q_8$ is solved by $S_3, S_5, S_{10}, S_{12}$; $Q_9$ is solved by $S_3, S_6, S_8, S_{13}$; $Q_{10}$ is solved by $S_3, S_7, S_9, S_{11}$; $Q_{11}$ is solved by $S_4, S_5, S_8, S_{11}$; $Q_{12}$ is solved by $S_4, S_6, S_9, S_{12}$; $Q_{13}$ is solved by $S_4, S_7, S_{10}, S_{13}$
satisfies the conditions of the question.