Example 4 Given that in a mathematics competition, the number of competition problems is $n(n \geqslant 4)$, each problem is solved by exactly 4 people, and for any two problems, there is exactly one person who solves both problems.

If the number of participants is no less than $4 n$, find the minimum value of $n$ such that there always exists a person who solves all the competition problems. (15th Korean Mathematical Olympiad)
Analysis and Solution First, when $4 \leqslant n \leqslant 13$, counterexamples can be constructed. In fact, because the number of participants $\geqslant 4 n \geqslant 16$, consider the following 13 sets: $M_{1}=\{1,2,3,4\}$, $M_{2}=\{1,5,6,7\}$, $M_{3}=\{1,8,9,10\}$, $M_{4}=\{1,11,12,13\}$, $M_{5}=\{2,5,8,11\}$, $M_{6}=\{2,6,9,12\}$, $M_{7}=\{2,7,10,13\}$, $M_{8}=\{3,5,9,13\}$, $M_{9}=\{3,6,10,11\}$, $M_{10}=\{3,7,8,12\}$, $M_{11}=\{4,5,10,12\}$, $M_{12}=\{4,7,9,11\}$, $M_{13}=\{4,6,8,13\}$. For $4 \leqslant n \leqslant 13$, take $n$ sets $M_{i}(1 \leqslant i \leqslant n)$, it is easy to verify that $\left|M_{i}\right|=4$, and $\left|M_{i} \cap M_{j}\right|=1$, but $M_{1}, \cdots, M_{n}$ have no common element, which does not meet the requirement. Therefore, we conjecture that $n_{\min }=14$.

Next, we only need to prove that $n=14$ satisfies the requirement. In fact, let the set of the numbers of people who solved the $i$-th problem be $M_{i}$, then for any $1 \leqslant i \leqslant 14$, $\left|M_{i}\right|=4$, and $\left|M_{i} \cap M_{i}\right|=1$. Let $M_{1}=\{a, b, c, d\}$, since each of the remaining 13 sets contains at least one of $a, b, c, d$, at least one, say $a$, appears at least 4 times in these 13 sets. Without loss of generality, assume $M_{2}$, $M_{3}$, $M_{4}$, $M_{5}$ all contain $a$, we will prove that every $M_{i}$ contains $a$.

By contradiction. Suppose there exists $t(6 \leqslant t \leqslant 14)$, such that $M_{t}$ does not contain $a$, but $M_{t}$ has one common element with $M_{1}$, $M_{2}$, $M_{3}$, $M_{4}$, $M_{5}$, let the common element of $M_{t}$ and $M_{k}(1 \leqslant k \leqslant 5)$ be $a_{k}(1 \leqslant k \leqslant 5)$, then $a_{k} \neq a$. If there exist $i, j(1 \leqslant i<j \leqslant 5)$, such that $a_{i}=a_{j}$, then $a, a_{i}$ both belong to $M_{i} \cap M_{j}$, which contradicts $\left|M_{i} \cap M_{j}\right|=1$. Therefore, $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ are all different, but they all belong to $M_{t}$, hence $\left|M_{t}\right| \geqslant 5$, which is a contradiction. Therefore, every $M_{t}$ contains $a$, thus the person $a$ solved all the problems, and the conclusion holds.

Hence, $n_{\min }=14$.