Problem 846

Let $n(x)$ denote the number of sets containing the element $x$. Let $T$ be the set of all elements in the 11 sets $M_{1}, M_{2}, \cdots, M_{11}$, i.e., $T=\bigcup_{i=1}^{11} M_{i}$.
From the problem, we know that $\sum_{x \in T} n(x)=5 \cdot 11=55$.
Let $C_{n(x)}^{2}$ denote the number of ways to choose two sets from $n(x)$ sets, which is also the number of ways to choose two sets containing the element $x$ from the 11 sets (if $n(x)=1$, we define $C_{n(x)}^{2}=0$). Since the intersection of any two sets $M_{i}, M_{j}(1 \leqslant i<j \leqslant 11)$ is non-empty, there is at least one common element $x \in T$ between $M_{i}$ and $M_{j}$. In other words, for any pair of sets $\left(M_{i}, M_{j}\right)(1 \leqslant i<j \leqslant 11)$, there is at least one common element $x \in T$. The pair $\left(M_{i}, M_{j}\right)$ is a pair of sets chosen from the $n(x)$ sets containing $x$, so we have
$\sum_{x \in T} C_{n(x)}^{2} \geqslant C_{11}^{2}=55$, i.e., $\frac{1}{2} \sum_{x \in T} n(x)[n(x)-1] \geqslant 55$.
Let $n=\max \{n(x) \mid x \in T\}$, then from the above inequality, we get $\frac{1}{2}(n-1) \sum_{x \in T} n(x) \geqslant 55$.
Using $(*)$, we get $\frac{1}{2}(n-1) \geqslant 1$, thus $n \geqslant 3$.
If $n=3$, then for any $x \in T$, we have $n(x) \leqslant 3$. We will prove that there does not exist $x \in T$ such that $n(x) \leqslant 2$.

By contradiction. If there exists some $x \in T$ such that $n(x) \leqslant 2$, then at least $11-2=9$ sets do not contain $x$. Without loss of generality, assume $M_{3}, M_{4}, \cdots, M_{11}$ do not contain $x$, and $M_{1}$ contains $x$. Since each of $M_{3}, M_{4}, \cdots, M_{11}$ has at least one common element with $M_{1}$, and this common element is not $x$, it must be one of the other 4 elements in $M_{1}$. These 4 elements belong to $M_{3}, M_{4}, \cdots, M_{11}$, which are 9 sets, so there must be an element $y$ that belongs to at least three of these 9 sets. Adding $y \in M_{1}$, we have $n(y) \geqslant 4$. This contradicts $n=3$.

Therefore, when $n=3$, for any $x \in T$, we have $n(x)=3$. Thus, we have $3 \cdot|T|=11 \cdot 5$, which gives $|T|=\frac{55}{3}$, a contradiction.
Hence, we get $n \geqslant 4$.
When $n=4$, we can provide an example that meets the conditions of the problem as follows:
$$\begin{array}{l} M_{1}=M_{2}=\{1,2,3,4,5\}, M_{3}=\{1,6,7,8,9\}, \\ M_{4}=\{1,10,11,12,13\}, M_{5}=\{2,6,9,10,14\}, \\ M_{6}=\{3,7,11,14,15\}, M_{7}=\{4,8,9,12,15\}, \\ M_{8}=\{5,9,13,14,15\}, M_{9}=\{4,5,6,11,14\}, \\ M_{10}=\{2,7,11,12,13\}, M_{11}=\{3,6,8,10,13\} . \end{array}$$