Example 15 (2000 National High School Competition Question) There are $n$ people, and it is known that any 2 of them make at most one phone call. The total number of calls made among any $n-2$ of them is equal, and is equal to $3^{k}$ ($k$ is a positive integer). Find all possible values of $n$.

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Let the total number of calls among $n$ people be $m$. Since $n$ people can form $C_{n}^{n-2}$ groups of $n-2$ people, and the total number of calls among each $n-2$ people is $3^{k}$, the total number of calls in all $n-2$ groups is $C_{n}^{n-2} \cdot 3^{k}$. On the other hand, in the above counting, each pair of people belongs to $C_{n-2}^{n-2}$ groups of $n-2$ people, so each call between two people is counted $C_{n-2}^{n-1}$ times. Therefore,
$$
m=\frac{C_{n}^{n-2} \cdot 3^{k}}{C_{n-2}^{n-1}}=\frac{n(n-1) \cdot 3^{k}}{(n-2)(n-3)} .
$$

And from $C_{n-2}^{2} \geqslant 3^{k} \geqslant 3$, we get $n \geqslant 5$.
(1) If 3 does not divide $n$, i.e., $(n, 3)=1$, then $(n-3, n)=1,\left(n-3,3^{k}\right)=1$, so $n-3 \mid n-1$, i.e., $\frac{n-1}{n-3}=1+\frac{2}{n-3}$ is a positive integer. Therefore, $n-3 \leqslant 2, n \leqslant 5$, and since $n \geqslant 5$, we have $n=5$.
(2) If 3 divides $n$, then $3 \mid n-3,3 \times n-2$, i.e., $(n-2,3)=1$. Also, $(n-2, n-1)=1$, so
$(n-2) \mid n$, i.e., $\frac{n}{n-2}=1+\frac{2}{n-2}$ is a positive integer.
Therefore, $n-2 \leqslant 2, n \leqslant 4$, which contradicts $n \geqslant 5$.
From (1) and (2), we know that $n$ can only be 5. On the other hand, if there are $n=5$ people, and each pair of them makes one call, then the total number of calls among any $n-2=3$ people is $C_{3}^{2}=3^{1}$ (here $k=1$, a positive integer), which meets the requirements of the problem. Therefore, the only positive integer solution is: $n=5$.