14. 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 of $k$ distinct 3-element subsets of $M$, where the intersection of any two 3-element subsets is non-empty.
14. When $n \leqslant 5$, it is obvious that $k=C_{n}^{3}$. Now assume $n \geqslant 6, M=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$, obviously, all $C_{n-1}^{2}$ three-element subsets of $M$ containing $a_{1}$ satisfy the problem's conditions, so $k \geqslant C_{n-1}^{2}$. On the other hand, let the maximum positive integer sought be $t$. If $A_{1}, A_{2}, \cdots, A_{t}$ are the three-element subsets that meet the conditions, consider the circular permutations of $a_{1}, a_{2}, \cdots, a_{n}$, which total $(n-1)!$ permutations, denoted as $\pi_{j}(j=1,2, \cdots,(n-1)!)$. If the three elements of subset $A_{i}$ are adjacent in the circular permutation $\pi_{j}$, then $A_{i}$ is said to belong to $\pi_{j}$, and the pair $\left(A_{i}, \pi_{j}\right)$ is formed. The total number of such pairs is denoted as $S$. Since each subset $A_{i}$ belongs to $3!(n-3)!$ circular permutations, and $i=1,2, \cdots, t$, we have $S=t \cdot 3!(n-3)!$. Furthermore, for each circular permutation $\pi_{j}$, when a three-element subset $A_{i}$ belongs to $\pi_{j}$, there can be at most two other three-element subsets that also belong to $\pi_{j}$ (in fact, if in $\pi_{j}$, $A_{i}=\left\{a_{i_{1}}, a_{i_{2}}, a_{i_{3}}\right\}$, and another three-element subset $A_{r}$ also belongs to $\pi_{j}$, then $A_{i}, A_{j}, A_{r}$ must have common elements pairwise, so it is only possible that $A_{i} \cap A_{i}=\left\{a_{i_{1}}, a_{i_{2}}\right\}$ and $A_{r} \cap A_{i}=\left\{a_{i_{2}}, a_{i_{3}}\right\}$, or $A_{j} \cap A_{i}=\left\{a_{i_{2}}, a_{i_{3}}\right\}$ and $A_{r} \cap A_{i}=\left\{a_{i_{1}}, a_{i_{2}}\right\}$, otherwise $A_{i} \cap A_{r}=\varnothing$). This indicates that for each $\pi_{j}$, there can be at most 3 three-element subsets that belong to $\pi_{j}$, meaning that the number of pairs containing $\pi_{j}$ is at most 3. Since $j=1,2, \cdots,(n-1)$, we have $S \leqslant 3 \cdot(n-1)!$, thus $t \cdot 3!(n-3)!\leqslant 3 \cdot(n-1)!$, leading to $t \leqslant \frac{3 \cdot(n-1)!}{3!(n-3)!}=\frac{(n-1)(n-2)}{2!}=C_{n-1}^{2}$. In summary, the maximum positive integer $k=C_{n}^{3}$ when $n \leqslant 5$ or $k=C_{n-1}^{2}$ when $n \geqslant 6$.