16. In a mathematics competition: (1) The number of competition problems is $n(n \geqslant 4)$; (2) Each problem is solved by exactly 4 people; (3) For any two problems, exactly 1 person solves both problems. If the number of participants is greater than or equal to $4 n$, find the minimum value of $n$ such that there always exists 1 person who solves all the competition problems.
(15th Korean Olympiad Problem)
16. Let $n \geqslant 14$, and associate each problem $p$ with a 4-element set $\{A, B, C, D\}$, where $A, B, C, D$ represent the contestants who solved the problem. From the given conditions, for any problem $p$, the 4-element sets corresponding to the other 13 problems all share a common element with the 4-element set corresponding to $p$. By the pigeonhole principle, there are at least $\left[\frac{13-1}{4}\right]+1=4$ problems $p_{1}, p_{2}, p_{3}, p_{4}$, whose 4-element sets share a common element with the 4-element set corresponding to $p$. Without loss of generality, assume the common element for $p, p_{1}, p_{2}; p_{3}, p_{4}$ is $A$. Let $p^{\prime}$ be any problem different from $p, p_{1}, p_{2}, p_{3}, p_{4}$. Then, the 4-element sets corresponding to $p_{0}, p_{1}, p_{2}, p_{3}, p_{4}$ each share a common element with the 4-element set corresponding to $p^{\prime}$. By the pigeonhole principle, there must be two problems $p_{i}, p_{j} (0 \leqslant i < j \leqslant 4)$ whose 4-element sets share the same common element $A^{\prime}$ with the 4-element set corresponding to $p^{\prime}$. If $A \neq A^{\prime}$, then $p_{i}$ and $p_{j}$ are solved by both $A$ and $A^{\prime}$, which contradicts the given condition (3). Therefore, $A = A^{\prime}$. By the arbitrariness of $p^{\prime}$, it follows that each problem's 4-element set has the same common element $A$, meaning $A$ solved all the problems. Thus, when $n \geqslant 14$, there is always a person who solved all the problems. The table below is an example where no one solved all 13 problems. The competition problems are numbered as $a_{1}, a_{2}, \cdots, a_{13}$, and the contestants are numbered as $1, 2, 3, \cdots$. In the table, except for contestants 1 to 13, no other contestants solved any problems (their numbers do not appear in the table). The first row represents the problem numbers, and the $j$-th column $(1 \leqslant j \leqslant 13)$ represents the numbers of the contestants who solved problem $a_{j}$ (in no particular order).

For $4 \leqslant n \leqslant 12$, removing some columns from the above table yields similar examples, so the minimum value of $n$ is 14.