Example 8 One evening, 21 people made $n$ phone calls to each other. It is known that among them, $m$ ($m$ is an odd number) people $a_{1}, a_{2}$, $\cdots, a_{m}$ made phone calls such that $a_{i}$ and $a_{i+1}\left(i=1,2, \cdots, m ; a_{m+1}=a_{1}\right)$ communicated. If among these 21 people, no three people communicated with each other, find the maximum value of $n$. ${ }^{[4]}$
【Analysis】Construct a graph theory model to characterize the relationships between elements in the problem. From the conditions given in the problem, we know that there is an odd cycle in the graph. Since the number of points is finite, by the extremal principle, there must be a shortest odd cycle $A_{1} A_{2} \cdots A_{2 k+1} A_{1}$. For any two points $A_{i}$ and $A_{j}(j \neq i \pm 1)$ within the shortest odd cycle, they must not be connected, thus the internal structure of the odd cycle is fully understood. At this point, starting from the shortest odd cycle $C$, the entire graph can be divided into two parts: one part is the odd cycle $A_{1} A_{2} \cdots A_{2 k+1} A_{1}$, and the other part is all the points except $A_{1} A_{2} \cdots A_{2 k+1} A_{1}$. We only need to understand the structure of each part and the connection situation between the two parts to grasp the overall structure.
Solution: Use 21 points to represent 21 people.
If two people communicate by phone, then connect the corresponding two points with an edge; otherwise, do not connect an edge, resulting in a graph $G$. It is known that graph $G$ contains an odd cycle, and there are no triangles in graph $G$, and we need to find the maximum value of the number of edges $n$ in graph $G$.

Assume the length of the shortest odd cycle $C$ in the graph is $2 k+1$.
Since there are no triangles in graph $G$, we have $k \geqslant 2$.
Let $C$ be $A_{1} A_{2} \cdots A_{2 k+1} A_{1}$. Then for any $i, j(1 \leqslant i, j \leqslant 2 k+1)$, $A_{i}$ and $A_{j}(j \neq i \pm 1)$ are not connected $\left(A_{0}=A_{2 k+2}=A_{1}\right)$ (otherwise, there would be a shorter odd cycle, contradicting the assumption). For the remaining
$$
21-(2 k+1)=2(10-k)
$$

points, there are no triangles.
By Turán's theorem, the maximum number of edges among these $2(10-k)$ points is $\frac{1}{4}(20-2 k)^{2}=(10-k)^{2}$. Additionally, each of these $2(10-k)$ points cannot be connected to two adjacent vertices in $C$ (otherwise, there would be a triangle, contradicting the assumption), so each of these $2(10-k)$ points can be connected to at most $k$ vertices in $C$. Therefore, the number of edges $n$ in the graph satisfies
$$
\begin{array}{l}
n \leqslant 2 k+1+(10-k)^{2}+2(10-k) k \\
=102-(k-1)^{2} \leqslant 102-(2-1)^{2}=101 .
\end{array}
$$

Suppose the 21 vertices of $G$ are $A_{1}, A_{2}, \cdots, A_{21}$,
$$
\begin{array}{l}
X=\left\{A_{1}, A_{2}, \cdots, A_{5}\right\}, \\
Y=\left\{A_{6}, A_{8}, A_{10}, \cdots, A_{20}\right\}, \\
Z=\left\{A_{7}, A_{9}, A_{11}, \cdots, A_{21}\right\} .
\end{array}
$$
In $X$, $A_{i}$ is connected to $A_{i+1}(i=1,2,3,4)$ and $A_{5}$ is connected to $A_{1}$, in $Y$ each point is connected to each point in $Z$, each point in $Y$ is connected to two points $A_{1}, A_{3}$ in $X$, and each point in $Z$ is connected to two points $A_{2}, A_{4}$ in $X$, and no other two points are connected. Then the number of edges in graph $G$ is
$$
n=5+8 \times 8+2 \times 8 \times 2=101 \text {, }
$$

and graph $G$ contains an odd cycle $A_{1} A_{2} \cdots A_{5} A_{1}$, but no triangles.
In summary, the maximum value of $n$ is 101.