The thirteenth question: One evening, 21 people made phone calls, and someone found that these 21 people made a total of $n$ calls, with each pair of people making at most one call. He also found that there were $\mathrm{m}$ people, the 1st person called the 2nd person, the 2nd person called the 3rd person, $\cdots$, the $\mathrm{m}-1$ person called the $\mathrm{m}$ person, and the $\mathrm{m}$ person called the 1st person. He refused to reveal the specific value of $\mathrm{m}$, only saying that $\mathrm{m}$ is an odd number. Try to find the minimum value of $n$ such that, regardless of the call situation, there must be 3 people among the 21 people who have called each other.
The thirteenth question:
Prove: Using 21 points to represent 21 people, if two people have communicated, then connect a line between the corresponding two points, thus obtaining a simple graph $\mathrm{G}(\mathrm{V}, \mathrm{E})$ of order 21, where $|\mathrm{V}|=21,|\mathrm{E}|=\mathrm{n}$. According to the problem, there exists a cycle of length $\mathrm{m}$ in graph $\mathrm{G}$. We need to find the minimum value of $\mathrm{n}$ such that if there is an odd cycle in $\mathrm{G}$, then there must be a triangle.
(1) First, prove: When $\mathrm{n}=102$, if $\mathrm{G}$ has an odd cycle, then it must have a triangle.
Let the shortest odd cycle in graph $\mathrm{G}$ be $\mathrm{C}$, with length $2 \mathrm{k}+1$.
If $k=1$, then $\mathrm{C}$ itself is a triangle, and the proposition is proved.
If $k \geq 2$, let $\mathrm{C}$ be $v_{1} v_{2} \ldots v_{2 k+1} v_{1}$, then no two non-adjacent vertices in $\mathrm{C}$ are connected. Otherwise, a new odd cycle with a length less than $\mathrm{C}$ would be generated, which contradicts the minimality of $\mathrm{C}$'s length.

Assume that the $21-(2 \mathrm{k}+1)=20-2 \mathrm{k}$ points outside $\mathrm{v}_{1}, \mathrm{v}_{2}, \ldots, \mathrm{v}_{2 \mathrm{k}+1}$ do not form a triangle. According to Turán's theorem, they can have at most $(10-\mathrm{k})^{2}$ edges. Furthermore, any of these points cannot be adjacent to two adjacent points in $\mathrm{C}$ (otherwise, a triangle would be formed), so each of these points can be adjacent to at most $\mathrm{k}$ points in $\mathrm{C}$. Therefore, the number of edges in graph $\mathrm{G}$ is no more than
$$
(2 \mathrm{k}+1)+(10-\mathrm{k})^{2}+(20-2 \mathrm{k}) \mathrm{k}=102-(\mathrm{k}-1)^{2}<102
$$

This contradicts the fact that graph $\mathrm{G}$ has 102 edges.
Therefore, when $\mathrm{n} \geq 102$, $\mathrm{G}$ must have a triangle.
(2) Below is a counterexample for $n=101$: Let the 21 vertices of $\mathrm{G}$ be $\mathrm{A}_{1}, \mathrm{A}_{2}, \ldots \mathrm{A}_{21}$, and define
$$
\begin{array}{l}
P=\left\{A_{1}, A_{2}, A_{3}, A_{4}, A_{5}\right\} ; \\
Q=\left\{A_{6}, A_{8}, A_{10}, \ldots, A_{20}\right\} ; \\
R=\left\{A_{7}, A_{9}, A_{11}, \ldots A_{21}\right\} \text { . } \\
\end{array}
$$

In point set $P$, $A_{i}$ and $A_{i+1}$ (for $i=1, 2, \ldots, 4$) are connected, and $A_{5}$ and $A_{1}$ are connected; each point in point set $Q$ is connected to each point in point set $R$; each point in point set $Q$ is connected to $A_{1}$ and $A_{3}$; each point in point set $R$ is connected to $A_{2}$ and $A_{4}$; no other points are connected. Then graph $\mathrm{G}$ has exactly $5+8 \times 8+2 \times 8+2 \times 8=101$ edges. At this point, graph $\mathrm{G}$ has an odd cycle $\mathrm{A}_{1} \mathrm{~A}_{2} \mathrm{~A}_{3} \mathrm{~A}_{4} \mathrm{~A}_{5} \mathrm{~A}_{1}$, but no triangle.
In conclusion, the minimum value of $n$ is 102.