Three, (50 points) $n$ people exchange greetings by phone during a holiday. It is known that each person makes calls to at most three friends, any two people make at most one call to each other, and among any three people, at least two of them, one person has called the other. Find the maximum value of $n$.


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.