Example 2 One evening, 21 people made $n$ phone calls to each other. It is known that among them, $m$ people $a_{1}, a_{2}, \cdots, a_{m}$ made phone calls to each other such that $a_{i}$ and $a_{i+1}$ communicated $\left(i=1,2, \cdots, m, a_{m+1}=a_{1}\right)$, and $m$ is odd. If among these 21 people, no 3 people made phone calls to each other, find the maximum value of $n$.
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, forming a graph $G$. It is known that $G$ contains an odd cycle, and there are no triangles in $G$. The task is to find the maximum number of edges $n$ in $G$.

Let the length of the shortest odd cycle $C$ in the graph be $2k+1$. Since there are no triangles in $G$, $k \geq 2$. Let $C$ be $A_{1} A_{2} \cdots A_{2k+1} A_{1}$. Then for any $i, j (1 \leq i, j \leq 2k+1)$, $A_{i}$ and $A_{j} (j \neq i \pm 1)$ are not connected $\left(A_{0} = A_{2k+2} = A_{1}\right)$ (otherwise, there would be a shorter odd cycle, contradicting the assumption). Apart from $A_{1}, A_{2}, \cdots, A_{2k+1}$, the remaining $21 - (2k+1) = 2(10-k)$ points have no triangles. By Turán's theorem, these $2(10-k)$ points can have at most $\frac{1}{4} (20-2k)^2 = (10-k)^2$ edges. 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 $C$. Thus, the number of edges $n$ in the graph satisfies:
$$
n \leq 2k+1 + (10-k)^2 + 2(10-k) \cdot k = 102 - (k-1)^2 \leq 102 - (2-1)^2 = 101.
$$

On the other hand, let the 21 vertices of $G$ be $A_{1}, A_{2}, \cdots, A_{21}$, $X = \{A_{1}, A_{2}, \cdots, A_{5}\}$, $Y = \{A_{6}, A_{8}, \cdots, A_{20}\}$, $Z = \{A_{7}, A_{9}, A_{11}, \cdots, A_{21}\}$. 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 points are connected. Then the number of edges in $G$ is:
$$
n = 5 + 8 \times 8 + 2 \times 8 \times 2 = 101.
$$

And $G$ 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.