Obviously, $n \geqslant 5$. Let the $n$ people be several points $A_{1}, A_{2}, \cdots, A_{n}$. If $A_{i}, A_{j}$ make a phone call, then connect $\left(A_{i}, A_{j}\right)$. Therefore, among these $n$ points, there must be an edge, let's say $\left(A_{1}, A_{2}\right)$.

Suppose $A_{1}$ and $A_{3}$ are not connected by an edge, and consider the $n-2$ points $A_{1}, A_{4}, A_{5}, \cdots, A_{n}; A_{2}, A_{4}, A_{5}, \cdots, A_{n}$; and $A_{3}, A_{4}, A_{5}, \cdots, A_{n}$. By the problem's condition, $A_{1}, A_{2}, A_{3}$ are each connected to $A_{4}, A_{5}, \cdots, A_{n}$ by the same number of edges, denoted as $k$.

Adding $A_{2}$ to the $n-1$ points $A_{1}, A_{4}, A_{5}, \cdots, A_{n}$, the total number of edges $S=3^{m}+k+1$. Removing any one point from these $n-1$ points, the remaining $n-2$ points are connected by $3^{m}$ edges, so each point is connected to the other $n-2$ points by $k+1$ edges. Thus, $S=\frac{1}{2}(n-1)(k+1)$.

Similarly, considering the $n-1$ points obtained by adding $A_{3}$ to $A_{1}, A_{4}, A_{5}, \cdots, A_{n}$, the total number of edges is $t=3^{m}+k=\frac{1}{2}(n-1) k$.

Since $S=t+1$, we get $\frac{1}{2}(n-1)(k+1)=\frac{1}{2}(n-1) k+1$, which simplifies to $n=3$, a contradiction. Therefore, $A_{1}$ and $A_{3}$ are connected by an edge.
Similarly, $A_{2}$ and $A_{3}$ are also connected by an edge. Thus, $A_{1}, A_{2}$ are connected to all $A_{i}(i=3,1, \cdots, n)$.

For $A_{i}, A_{j}(i \neq j)$, since $A_{i}$ is connected to $A_{1}$, it follows that $A_{i}$ is connected to $A_{j}$. Therefore, these $n$ points form a complete graph. So $3^{m}=\frac{1}{2}(n-2)(n-3)$. Hence, $n=5$.
There are $n$ people, and it is known that any two of them make at most one phone call. The total number of phone calls made among any $n-2$ of them is equal, and it is $3^{m}$ times, where $m$ is a natural number. Find all possible values of $n$.

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