Example 3: 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 $n-2$ people is equal, and it is $3^{k}$ times, where $k$ is a positive integer. Find all possible values of $n$.
(2000, National High School Mathematics Competition)
Analysis: Carefully savor the quantitative relationships given in the problem, and you can obtain its quantitative characteristics.

Assume the $n$ people are $A_{1}, A_{2}, \cdots, A_{n}$, and the number of calls made by $A_{i}$ is $m_{i}$. The number of calls between $A_{i}$ and $A_{j}$ is $\lambda_{i j}(1 \leqslant i, j \leqslant n)$, where $\lambda_{i j}=0$ or 1. Thus, the quantitative characteristic is
$$
m_{i}+m_{j}-\lambda_{i j}=\frac{1}{2} \sum_{s=1}^{n} m_{s}-3^{k}=c,
$$

where $c$ is a constant. Consequently, we conjecture that $m_{i}(i=1,2, \cdots, n)$ is a constant.
Let $n$ people be denoted as $A_{1}, A_{2}, \cdots, A_{n}$. Let the number of calls made by $A_{i}$ be $m_{i}$, and the number of calls between $A_{i}$ and $A_{j}$ be $\lambda_{i j}(1 \leqslant i, j \leqslant n)$, where $\lambda_{i j}=0$ or 1.
Clearly, $n \geqslant 5$. Therefore,
$$
\begin{array}{l}
\left|m_{i}-m_{j}\right|=\left|\left(m_{i}+m_{s}\right)-\left(m_{j}+m_{s}\right)\right| \\
=\left|\lambda_{i s}-\lambda_{j}\right| \leqslant 1,1 \leqslant i, j, s \leqslant n .
\end{array}
$$

Let $m_{i}=\max \left\{m_{s}, 1 \leqslant s \leqslant n\right\}$,
$$
m_{j}=\min \left\{m_{s}, 1 \leqslant s \leqslant n\right\} .
$$

Thus, $m_{i}-m_{j} \leqslant 1$.
If $m_{i}-m_{j}=1$, then for any $s \neq i, j, 1 \leqslant s \leqslant n$, we have
$$
\begin{array}{l}
\left(m_{i}+m_{s}-\lambda_{i s}\right)-\left(m_{j}+m_{s}-\lambda_{j s}\right) \\
=1-\left(\lambda_{i s}-\lambda_{j s}\right)=0,
\end{array}
$$

which means $\lambda_{i s}-\lambda_{j s} \equiv 1,1 \leqslant i, j, s \leqslant n$.
Therefore, $\lambda_{i s} \equiv 1, \lambda_{j i} \equiv 0, s \neq i, j$, and $1 \leqslant s \leqslant n$.
Thus, $m_{i} \geqslant n-2, m_{j} \leqslant 1$.
Hence, $m_{i}-m_{j} \geqslant n-3 \geqslant 2$, which is a contradiction.
Therefore, $m_{i}-m_{j}=0$.
Then $\lambda_{i j} \equiv 0$ or $\lambda_{i j} \equiv 1$.
If $\lambda_{i j} \equiv 0$, then $m_{s} \equiv 0,1 \leqslant s \leqslant n$, which is a contradiction.
If $\lambda_{i j} \equiv 1$, then $m_{s}=n-1,1 \leqslant s \leqslant n$.
Thus, $(n-2)(n-3)=3^{k} \times 2$.
Let $n-2=2 \times 3^{k_{1}}, n-3=3^{k_{2}}, k_{1} \geqslant k_{2}$.

From $2 \times 3^{k_{1}}-3^{k_{2}}=1$, we get
$$
3^{k_{2}}=1 \text{. }
$$

Therefore, $k_{1}=k_{2}=0$. This contradicts $k \geqslant 1$.
Let $n-2=3^{k_{1}}, n-3=2 \times 3^{k_{2}}, k_{1} \geqslant k_{2}+1$.
From $3^{k_{1}}-2 \times 3^{k_{2}}=1$, we get
$$
k_{2}=0, k_{1}=1 \text{. }
$$

Therefore, $n=5$ is the solution.