13. In a mathematics competition,
(i) the number of problems is $n(n \geqslant 4)$;
(ii) each problem is solved by exactly 4 people;
(iii) 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 a person who solves all the competition problems.
(15th Korean Mathematical Olympiad)
Solution: Let $n \geqslant 14$, and represent each problem as a rectangle, with its 4 vertices representing the 4 contestants who solved the problem. Let $P$ be any problem.

Each problem different from $P$ shares a common vertex with $P$, since $n \geqslant 14$. By the pigeonhole principle, there is a vertex of $P$, denoted as $S$, that is a vertex of at least 4 (different from $P$) rectangles.

Assume there exists a rectangle $P$ that does not share vertex $S$ with $P$, then there are at least 5 rectangles with vertex $S$, and each of these rectangles shares only one common vertex with $P$. By the pigeonhole principle, a vertex $S^{\prime}$ of $P$ is a vertex of at least 2 rectangles that have $S$ as a vertex.

If $S \neq S^{\prime}$, then there are at least 2 rectangles that share both vertex $S$ of $P$ and vertex $S^{\prime}$ of $P$, which contradicts condition (iii).
Hence, $S=S^{\prime}$, meaning all rectangles share vertex $S$.
Table 1 is an example when $n=13$ where no one solved all the problems. Number the problems as $1,2, \cdots, 13$, and number the contestants as 1, $2, \cdots$. In Table 1, all contestants except for contestants 1 to 13 did not solve any problems. The first row represents the problem numbers, and the $i$-th column $(1 \leqslant i \leqslant 13)$ represents the numbers of the 4 contestants who solved the $i$-th problem.
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline Competition & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 \\
\hline \multirow{4}{*}{\begin{tabular}{l} 
Contestants who \\ 
solved the problem
\end{tabular}} & 1 & 1 & 1 & 1 & 2 & 2 & 2 & 3 & 3 & 3 & 4 & 4 & 4 \\
\hline & $?$ & 5 & 8 & 11 & 5 & 6 & 7 & 5 & 6 & 7 & 5 & 6 & 7 \\
\hline & 5 & 6 & 9 & 12 & 8 & 9 & 10 & 9 & 10 & 8 & 10 & 8 & 9 \\
\hline & 4 & 7 & 10 & 13 & 11 & 12 & 13 & 13 & 11 & 12 & 12 & 13 & 11 \\
\hline
\end{tabular}

For $4 \leqslant n \leqslant 12$, similar examples can be obtained by removing some columns (problems). Therefore, the minimum value is 14.