Third question: There are $n$ people, and it is known that any two of them make at most one phone call. 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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Solution 1: Clearly, $n \geqslant 5$. Let the $n$ people be $n$ points $A_{1}, A_{2}, \cdots, A_{n}$. If $A_{i}$ and $A_{j}$ make a phone call, then connect $A_{i} A_{j}$. Therefore, there must be line segments among these $n$ points. Without loss of generality, assume there is a line segment between $A_{1} A_{2}$:

If there is no line segment between $A_{1} A_{3}$, 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, the total number of line segments between $A_{1}$, $A_{2}$, and $A_{3}$ and $A_{4}$, $A_{5}$, $\cdots$, $A_{n}$ are all equal, denoted as $m$.
Adding $A_{2}$ to $A_{1} A_{4} A_{5} \cdots A_{n}$, the total number of line segments among these $n-1$ points is $s=3^{k}+m+1$. Removing any one point from these $n-1$ points, the total number of line segments among the remaining $n-2$ points is $3^{k}$, so each point is connected to the other $n-2$ points by $m+1$ line segments. Thus,
$$
s=\frac{1}{2}(n-1)(m+1).
$$

Adding $A_{3}$ to $A_{1} A_{4} A_{5} \cdots A_{n}$, the total number of line segments among these $n-1$ points is $t=3^{k}+m$. Similarly, we get $t=\frac{1}{2}(n-1) m$.
Therefore, from $s=t+1$ we have
$\frac{1}{2}(n-1)(m+1)=\frac{1}{2}(n-1) m+1$, which implies $n=3$, a contradiction. So there must be a line segment between $A_{1} A_{3}$.

Similarly, there must be a line segment between $A_{2} A_{3}$. Therefore, $A_{1}$ and $A_{2}$ are connected to all $A_{i}(i=3,4, \cdots, n)$.

For $A_{i}$ and $A_{j}(i \neq j)$, since $A_{i}$ is connected to $A_{1}$, similarly $A_{i}$ is connected to $A_{j}$. Therefore, $A_{i}$ and $A_{j}$ $(i, j=1,2, \cdots, n)$ are all connected.
Thus, $3^{k}=\frac{1}{2}(n-2)(n-3)$. Hence, $n=5$.
(Provided by Class 1, Senior 3, Xinhua High School, Tianjin, 300201, Tutor: Wang Hao)

Solution 2: Clearly, $n \geqslant 5$. Let the $n$ people be $A_{1}$, $A_{2}, \cdots, A_{n}$. Let the number of calls made by $A_{i}$ be $d\left(A_{i}\right), i=1,2, \cdots, n$.

If $d\left(A_{1}\right)=0$, among $A_{3}, A_{4}, \cdots, A_{n}$, there must be one person, say $A_{3}$, who has called at least one of $A_{4}$, $A_{5}, \cdots, A_{n}$. Thus, the total number of calls between $A_{1}, A_{4}, A_{5}, \cdots, A_{n}$ and $A_{3}, A_{4}, \cdots, A_{n}$ are not equal, a contradiction. Therefore, $d\left(A_{i}\right) \neq 0(i=1,2, \cdots, n)$.

If $d\left(A_{1}\right) \neq d\left(A_{2}\right)$, assume without loss of generality that $d\left(A_{1}\right)<d\left(A_{2}\right)$. Among $A_{3}, A_{4}, \cdots, A_{n}$, if there is one person, say $A_{3}$, who has called both $A_{1}$ and $A_{2}$ or neither, then the total number of calls between $A_{1}, A_{4}, A_{5}, \cdots, A_{n}$ and $A_{2}, A_{4}, A_{5}, \cdots, A_{n}$ are not equal, a contradiction. If each person in $A_{3}, A_{4}, \cdots, A_{n}$ either calls $A_{1}$ but not $A_{2}$ or calls $A_{2}$ but not $A_{1}$, when all $A_{3}, A_{4}, \cdots, A_{n}$ call $A_{2}$, $A_{1}$ can only call $A_{2}$, so the total number of calls between $A_{1}, A_{4}, A_{5}, \cdots, A_{n}$ is less than that between $A_{2}, A_{4}, \cdots, A_{n}$. When there is one person, say $A_{3}$, who calls $A_{1}$ but not $A_{2}$, the total number of calls between $A_{1}, A_{4}, \cdots, A_{n}$ is still less than that between $A_{2}, A_{4}, A_{5}, \cdots, A_{n}$, a contradiction. Therefore, $d\left(A_{i}\right)=d\left(A_{j}\right), i \neq j, i, j=1,2, \cdots, n$. Let this common value be $d$.
If $1 \leqslant a \leqslant n-2$, then there exist $A_{i}, A_{j}(i \neq j)$ who have not called each other, and there exist $A_{s}, A_{t}(s \neq t)$ who have called each other. The total number of calls among the $n-2$ people excluding $A_{i}, A_{j}$ is $\frac{1}{2} n d-2 d$, and the total number of calls among the $n-2$ people excluding $A_{s}, A_{t}$ is $\frac{1}{2} n d-2 d+1$, a contradiction. Therefore, $d=n-1$.
Thus, $\frac{1}{2} n(n-1)-2(n-1)+1=3^{k}$, i.e., $(n-2)(n-3)=2 \times 3^{k}$. Hence, $n=5$.
(Provided by Zhu Hengjie, Zibo Teaching Research Office, Shandong Province, 255033)

Solution 3: Clearly, $n \geqslant 5$. Since $n$ people can form $\mathrm{C}_{n}^{n-2}=\mathrm{C}_{n}^{2}$ "groups of $n-2$ people", the total number of calls among $n$ people, counted with multiplicity, is $3^{k} C_{n}^{2}$.
If two people make a call, this call is counted once in $\mathrm{C}_{n-2}^{n-4}=\mathrm{C}_{n-2}^{2}$ "groups of $n-2$ people". Therefore, the multiplicity of each call is $\mathrm{C}_{n-2}^{2}$. Hence, the actual total number of calls among $n$ people is $l=\frac{C_{n}^{2} 3^{k}}{\mathrm{C}_{n-2}^{2}}$, where $l$ is a natural number, i.e.,
$$
\frac{(n-2)(n-3)}{2} l=\frac{n(n-1)}{2} 3^{k}.
$$

Since one of $n$ and $n-2$ or $n-1$ and $n-3$ is a multiple of 4 and the other is an even number but not a multiple of 4, $\frac{1}{2}(n-2)(n-3)$ and $\frac{1}{2} n(n-1)$ are one odd and one even.
When $n_{1}$ is odd, $\frac{n_{1}-1}{2}=1, 3 n_{1}-2=7$, $7 \mid 3^{k}$, a contradiction. When $n_{1}$ is even, $n_{1}-1=1$, $\frac{1}{2}\left(3 n_{1}-2\right)=2,2 \mid 3^{k}$, a contradiction. Hence, $d=1$, and we have
$$
\left.\frac{1}{2}(n-2)(n-3) \right\rvert\, 3^{k}.
$$

If $n$ is even, since $\left(\frac{n-2}{2}, n-3\right)=1$, we have $\frac{n-2}{2}=1, n=4$, a contradiction.
If $n$ is odd, we have $\frac{n-3}{2}=1, n=5$.
When $n=5$, the condition is satisfied.