Example 8 One evening, 21 people made $n$ phone calls to each other. It is known that among them, there are $m$ ($m$ is an odd number) people $a_{1}, a_{2}$, $\cdots, a_{m}$ such that $a_{i}$ and $a_{i+1}\left(i=1,2, \cdots, m ; a_{m+1}=a_{1}\right)$ made phone calls. If no three people among these 21 people made phone calls to each other, find the maximum value of $n$.

【Analysis】Construct a graph theory model to characterize the relationships between the elements in the problem. From the conditions given in the problem, we know that there exists an odd cycle in the graph. Since the number of points is finite, by the extreme principle, we know that there must exist the shortest odd cycle $A_{1} A_{2} \cdots A_{2 k+1} A_{1}$. Furthermore, in the shortest odd cycle, any two points $A_{i}$ and $A_{j}(j \neq i \pm 1)$ 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.
Solve: Represent 21 people with 21 points.
If two people communicate by phone, 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 $G$. The task is to find the maximum number of edges $n$ in graph $G$.

Let the length of the shortest odd cycle $C$ in the graph be $2k+1$.
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 smaller odd cycle, contradicting the assumption). Apart from $A_{1}, A_{2}, \cdots, A_{2 k+1}$,

the remaining
$$
21-(2 k+1)=2(10-k)
$$

points have 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). Therefore, each of these $2(10-k)$ points can be connected to at most $k$ vertices in $C$. Hence, 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 $Z$, and each point in $Y$ is connected to two points in $X$, $A_{1}$ and $A_{3}$. Each point in $Z$ is connected to two points in $X$, $A_{2}$ and $A_{4}$. No other pairs of 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 conclusion, the maximum value of $n$ is 101.