2. 7511 sets $M_{1}, M_{2}, \cdots, M_{11}$. Each set has 5 elements, and the intersection of any two sets is non-empty. Find the minimum possible value of the maximum number of sets that have a common element.
[Solution] 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.,
$$
T=\bigcup_{i=1}^{11} M_{i} \text {. }
$$

From the problem, we know that,
$$
\sum_{x \in T} n(x)=5 \cdot 11=55.
$$
$C_{n(x)}^{2}$ is the number of pairs of sets that can be formed by choosing any two sets from the $n(x)$ sets, which is also the number of pairs of sets that can be formed by choosing any two sets from the 11 sets that contain the element $x$ (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, any pair of sets $(M_{i}, M_{j})(1 \leqslant i<j \leqslant 11)$ has at least one common element $x \in T$. $(M_{i}, M_{j})$ is a pair of sets from the $n(x)$ sets containing $x$. Therefore, we have
$$
\sum_{x \in T} C_{n(x)}^{2} \geqslant C_{11}^{2}=55 \text {, }
$$

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 (1), we get
$$
\frac{1}{2}(n-1) \geqslant 1 \text {, }
$$

thus
$$
n \geqslant 3 \text {. }
$$

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 there are at least $11-$ $2=9$ sets that 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 the sets $M_{3}, M_{4}, \cdots, M_{11}$, plus $y \in M_{1}$, 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
$$
\begin{aligned}
3 \cdot|T| & =11 \cdot 5 \\
|T| & =\frac{55}{3}
\end{aligned}
$$

Contradiction.
Therefore, $n \geqslant 4$.
When $n=4$, we can provide an example that meets the conditions of the problem as follows:
$$
\begin{array}{l}
M_{1}=M_{2}=\{1,2,3,4,5\} \\
M_{3}=\{1,6,7,8,9\} \\
M_{4}=\{1,10,11,12,13\} \\
M_{5}=\{2,6,9,10,14\} \\
M_{6}=\{3,7,11,14,15\} \\
M_{7}=\{4,8,9,12,15\} \\
M_{8}=\{5,9,13,14,15\} \\
M_{9}=\{4,5,6,11,14\} \\
M_{10}=\{2,7,11,12,13\} \\
M_{11}=\{3,6,8,10,13\}
\end{array}
$$