$7 \cdot 67$ Given 11 sets $M_{1}, M_{2}, \cdots, M_{11}$, where each set has exactly 5 elements and for all $i, j, 1 \leqslant i<j \leqslant 11$, there is $M_{i} \cap M_{j} \neq \varnothing$, find the minimum possible value of the maximum number of sets among these sets whose intersection is non-empty.
[Solution] Consider the following 11 sets:
$$
\begin{array}{ll}
\{1,2,3,4,5\}, & \{1,6,7,8,9\}, \\
\{2,6,10,11,12\}, & \{3,7,10,13,14\}, \\
\{4,8,11,13,15\}, & \{5,9,12,14,15\}, \\
\{1,10,15,16,17\}, & \{2,8,14,16,18\}, \\
\{3,9,11,17,18\}, & \{4,7,12,16,18\}, \\
\{5,6,13,17,18\}, &
\end{array}
$$

It can be seen that they satisfy the requirements of the problem, and except for 18 appearing 4 times, each number appears exactly 3 times. Therefore, the minimum possible value is no greater than 4.

On the other hand, the 11 sets contain a total of 55 elements. If the number of distinct elements is no more than 18, then by the pigeonhole principle, there must be an element that belongs to at least 4 sets, so the minimum possible value is at least 4. If the number of distinct elements in the 11 sets is at least 19, then by the pigeonhole principle, there must be an element that belongs to at most 2 sets. Suppose this element is in $\{1,2,3,4,5\}$ and $\{1, \times, \times, \times, \times\}$, and the other sets do not contain 1. Then the other 9 sets must each contain at least one of $\{2,3,4,5\}$. Thus, by the pigeonhole principle, at least 3 sets must contain the same element from $\{2,3,4,5\}$, which again leads to the minimum possible value being at least 4.
In summary, the minimum possible value of the maximum number of sets with non-empty intersection is 4.