Problem 722

Three, first prove the lemma.
Lemma In a simple graph $G$ of order $n$ without $K_{3}$, then
$$f(G) \leqslant\left[\frac{n^{2}}{4}\right],$$

where $f(G)$ represents the number of edges in $G$.
Proof of the lemma: Let $A$ be the vertex with the maximum degree, and let the set of vertices adjacent to $A$ be $M=\left\{A_{1}, A_{2}, \cdots, A_{r}\right\}$, and the set of vertices not adjacent to $A$ be $N=\left\{B_{1}, B_{2}, \cdots, B_{s}\right\}(r+s+1=n)$. Since $G$ contains no triangles, there are no edges in $M$. Thus, the other edges of $G$ are either in $N$ or between $M$ and $N$. These edges are all induced by the vertices $B_{1}, B_{2}, \cdots, B_{s}$. Therefore,
$$\begin{array}{l} f(G) \leqslant d(A)+d\left(B_{1}\right)+d\left(B_{2}\right)+\cdots+d\left(B_{s}\right) \\ \leqslant r+\underbrace{r+\cdots+r}_{s}=(s+1) r \leqslant\left(\frac{s+r+1}{2}\right)^{2}=\frac{n^{2}}{4} . \end{array}$$

Since $f(G) \in \mathbf{Z}$, we have
$$f(G) \leqslant\left[\frac{n^{2}}{4}\right] \text {. }$$

Now, prove the original problem.
Use $n$ points to represent $n$ people. If a person $A$ calls another person $B$'s home phone, then draw a directed edge from $A$ to $B$, resulting in a simple directed graph $G$.
On one hand, $\bar{G}$ contains no triangles. By the lemma,
$f(\bar{G}) \leqslant\left[\frac{n^{2}}{4}\right]$.
Thus, $f(G)=\mathrm{C}_{n}^{2}-f(\bar{G})$
$\geqslant C_{n}^{2}-\left[\frac{n^{2}}{4}\right]=\left[\frac{(n-1)^{2}}{4}\right]$.
On the other hand, $f(G)=\sum_{i=1}^{n} d^{+}\left(x_{i}\right) \leqslant \sum_{i=1}^{n} 3=3 n$.
Therefore, $\left[\frac{(n-1)^{2}}{4}\right] \leqslant 3 n$.
When $n$ is odd, equation (1) becomes $\frac{(n-1)^{2}}{4} \leqslant 3 n$, solving for $n \leqslant 13$;
When $n$ is even, equation (1) becomes $\frac{n^{2}-2 n}{4} \leqslant 3 n$, solving for $n \leqslant 14$.
In summary, $n \leqslant 14$.
Finally, $n=14$ is possible. Construct two $K_{7}$, for each heptagon $A_{1} A_{2} \cdots A_{7}$, let $A_{i}$ point to $A_{i+1}, A_{i+2}, A_{i+3}(i=1,2, \cdots, 7, A_{i+7}=A_{i})$, then the graph meets the conditions.

First, each vertex as a starting point exactly induces 3 directed edges, thus, each person calls at most 3 friends' home phones.

Second, for any three vertices, by the pigeonhole principle, there must be two vertices $A_{i} 、 A_{j}(i<j)$ such that $A_{j}$ calls $A_{i}$'s home phone.