Three, there are $n$ people, it is known that any two of them make at most one phone call, and any $n-2$ of them have the same total number of phone calls, which is $3^{k}$ times, where $k$ is a natural number. Find all possible values of $n$.

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5
Three, 【Analysis and Solution】Obviously, $n \geqslant 5$. Let's assume the $n$ people are $A_{1}, A_{2}, \cdots, A_{n}$.
Also, let the number of calls made by $A_{i}$ be $m_{i}$, and the number of calls between $A_{i}$ and $A_{j}$ be $\lambda_{i, j}, 1 \leqslant i, j \leqslant n$. We have $m_{i}+m_{j}-\lambda_{i, j}=$ $\frac{1}{2} \sum_{s=1}^{n} m_{s}-3^{k}=c$.
(1) $c$ is a constant, and $1 \leqslant i, j \leqslant n$.
From the above equation, we know $\left|m_{i}-m_{j}\right|=\left|\left(m_{i}+m_{s}\right)-\left(m_{j}+m_{s}\right)\right|=\left|\lambda_{i, s}-\lambda_{j, s}\right| \leqslant 1,1 \leqslant i, j, s \leqslant n$.
That is, $\left|m_{i}-m_{j}\right| \leqslant 1 \quad 1 \leqslant i, j \leqslant n$.
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 $m_{i}-m_{j} \leqslant 1$.
When $m_{i}-m_{j}=1$, for any $s \neq i, j, 1 \leqslant s \leqslant n$, we have
$\left(m_{i}+m_{s}-\lambda_{i, s}\right)-\left(m_{j}+m_{s}-\lambda_{j, s}\right)=1-\left(\lambda_{i, s}-\lambda_{j, s}\right) \equiv 0$. That is, $\lambda_{i, s}-\lambda_{j, s} \equiv 1$.
Thus, $\lambda_{i, s} \equiv 1, \lambda_{j, s} \equiv 0 . s \neq i, j, 1 \leqslant s \leqslant n$,
$\therefore m_{i} \geqslant n-2, m_{j} \leqslant 1$, which implies $m_{i}-m_{j} \geqslant n-2-1 \geqslant 2$.
This is a contradiction, so $m_{i}-m_{j}=0$, meaning $m_{s}(1 \leqslant s \leqslant n)$ is always a constant.
From (1), we know $\lambda_{i, j} \equiv 0$ or $\lambda_{i, j} \equiv 1$.
When $\lambda_{i, j} \equiv 0$, we have $m_{s} \equiv 0,1 \leqslant s \leqslant n$, which contradicts the given conditions.
$$
\therefore \frac{1}{2} n(n-1)-(2 n-3)=3^{k} \text {, }
$$

That is, $(n-2)(n-3)=2 \times 3^{k}$.
Let $n-2=2 \times 3^{k_{1}}, n-3=3^{k_{2}}, k_{1} \geqslant k_{2}$,
Then $2 \times 3^{k_{1}}-3^{k_{2}}=1$,
$$
\therefore 3^{k_{2}}\left(2 \times 3^{k_{1}-k_{2}}-1\right)=1 \text {, }
$$

Thus, $3^{k_{2}}=1,2 \times 3^{k_{1}-k_{2}}-1=1 \Rightarrow k_{1}=0, k_{2}=0$.
But this contradicts $k \geqslant 1$.
$\therefore$ Let $n-2=3^{k_{1}}, n-3=2 \times 3^{k_{2}}, k_{1} \geqslant k_{2}+1$,
Then $3^{k_{1}}-2 \times 3^{k_{2}}=1$,
Similarly, we have $k_{1}=1, k_{2}=0, n=5$.
When each of the 5 people calls each other once, the total number of calls among any 3 people is $3^{1}$ times. Therefore, $n=$ 5 is the solution.