Three. (Full marks 50 points) 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 is $3^{k}$ times, where $k$ is a natural number. Find all possible values of $n$.

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Obviously, $n \geqslant 5$. Let the $n$ people be $A_{1}, A_{2}, \cdots, A_{n}$.
Suppose the number of calls made by $A_{i}$ is $m_{i}$, and the number of calls between $A_{i}$ and $A_{j}$ is $\lambda_{i, j}, 1 \leqslant i, j \leqslant n$. Then
$$
m_{i}+m_{j}-\lambda_{i, j}=\frac{1}{2} \sum_{s=1}^{n} m_{s}-3^{k}=c .
$$

where $c$ is a constant, $1 \leqslant i, j \leqslant n$.
According to $(*)$ we have
$$
\begin{array}{l}
\left|m_{i}-m_{j}\right|=\left|\left(m_{i}+m_{s}\right)-\left(m_{j}+m_{s}\right)\right| \\
=\left|\lambda_{i, s}-\lambda_{j, s}\right| \leqslant 1,1 \leqslant i, j, s \leqslant n \\
\Rightarrow\left|m_{i}-m_{j}\right| \leqslant 1,1 \leqslant i, j \leqslant n .
\end{array}
$$

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

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

i.e., $\lambda_{i, s} \equiv 1, \lambda_{j, s} \equiv 0, s \neq i, j, 1 \leqslant s \leqslant n$.
Therefore, $m_{i} \geqslant n-2, m_{j} \leqslant 1$.
Thus, $m_{i}-m_{j} \geqslant n-3 \geqslant 2$.
A contradiction arises, so $m_{i}-m_{j}=0$, i.e., $m_{s}(1 \leqslant s \leqslant n)$ is always a constant.
According to $(*)$, $\lambda_{i, j} \equiv 0$ or $\lambda_{i, j} \equiv 1$.
If $\lambda_{i, j} \equiv 0$, then $m_{s} \equiv 0,1 \leqslant s \leqslant n$. This contradicts the given information.
Therefore, $\lambda_{i, j} \equiv 1 \Rightarrow m_{s} \equiv n-1,1 \leqslant s \leqslant n$. So,
$$
\frac{1}{2} n(n-1)-(2 n-3)=3^{k} \text {, }
$$

i.e., $(n-2)(n-3)=2 \times 3^{k}$.
Let $n-2=2 \times 3^{k_{1}}, n-3=3^{k_{2}}, k_{1} \geqslant k_{2}$, then $2 \times 3^{k_{1}}-3^{k_{2}}=1$.
Thus, $3^{k_{2}}\left(2 \times 3^{k_{1}-k_{2}}-1\right)=1$.

We get $3^{k_{2}}=1,2 \times 3^{k_{1}-k_{2}}-1=1$.
Therefore, $k_{2}=0, k_{1}=0$. This contradicts $k \geqslant 1$.
Let $n-2=3^{k_{1}}, n-3=2 \times 3^{k_{2}}, k_{1} \geqslant k_{2}+1$, then $3^{k_{1}}-2 \times 3^{k_{2}}=1$.
Thus, $3^{k_{2}}\left(3^{k_{1}-k_{2}}-2\right)=1$.

We get $3^{k_{2}}=1,3^{k_{1}-k_{2}}-2=1$.
Therefore, $k_{2}=0, k_{1}=1, n=5$.
In this case, if each pair of the 5 people calls each other once, then the total number of calls among any 3 people is $3^{1}$ times.
In conclusion, $n=5$ is the only possible value for $n$.
(Provided by Lian Xiao)