Example 5 Let $n(\geqslant 3)$ be a positive integer, and $M$ be an $n$-element set. Find the maximum positive integer $k$ such that:
there exists a family $\psi$ of $k$ three-element subsets of $M$ such that the intersection of any two elements of $\psi$ (note: elements of $\psi$ are three-element sets) is non-empty.
When $n \leqslant 5$, it is obvious that the maximum positive integer $k=C_{n}^{3}$ satisfies the condition.
Below, we prove that when $n \geqslant 6$, the maximum positive integer $k=C_{n-1}^{2}$ satisfies the condition.
On one hand, let $M=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$. Then, the family of all three-element subsets of $M$ that include $a_{1}$ satisfies the condition, so the positive integer $k \geqslant C_{n-1}^{2}$.
On the other hand, let the maximum integer sought be $t$.
We consider the circular permutations of $a_{1}, a_{2}, \cdots, a_{n}$, then the total number of circular permutations is $(n-1)!$.
Consider the number of circular permutations in which the three elements of some $A_{i}$ in $\psi$ are adjacent, which is $x$, then $x=3!\cdot(n-3)!$.
Notice that when the three elements of $A_{i}$ are adjacent in the circular permutation, there can be at most two other elements $A_{j}$ and $A_{r}$ in $\psi$ whose elements are also adjacent in this circular permutation.
In fact, if $A_{i}=\left\{a_{i}, a_{i,}, a_{i,}\right\}$, then when the elements of $A_{j}$ and $A_{r}$ are also adjacent, it can only be that $A_{j} \cap A_{i}=\{a_{i,}, a_{i,}\}, A_{r} \cap A_{i}=\left\{a_{i_{1}}, a_{i_{1}}\right\}$, or their symmetric forms. Otherwise, $A_{i} \cap A_{r}=\varnothing$, which contradicts the problem's condition.
Therefore, there cannot be another element $A_{m}$ in $\psi$ such that the elements of $A_{m}$ are also adjacent in this circular permutation.
The above discussion shows that $t \cdot x \leqslant 3 \cdot((n-1)!)$,
Thus, $t \leqslant \frac{(n-1)!}{2!(n-3)!}=C_{n-1}^{2}$.