15th 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.