4. Consider any vertex $v_{n}$ from which the maximal number $d$ of segments start, and suppose it is not a vertex of a triangle. Let $\mathcal{A}=$ $\left\{v_{1}, v_{2}, \ldots, v_{d}\right\}$ be the set of points that are connected to $v_{n}$, and let $\mathcal{B}=\left\{v_{d+1}, v_{d+2}, \ldots, v_{n}\right\}$ be the set of the other points. Since $v_{n}$ is not a vertex of a triangle, there is no segment both of whose vertices lie in $\mathcal{A}$; i.e., each segment has an end in $\mathcal{B}$. Thus, if $d_{j}$ denotes the number of segments at $v_{j}$ and $m$ denotes the total number of segments, we have $$m \leq d_{d+1}+d_{d+2}+\cdots+d_{n} \leq d(n-d) \leq\left[\frac{n^{2}}{4}\right]=m$$ This means that each inequality must be equality, implying that each point in $\mathcal{B}$ is a vertex of $d$ segments, and each of these segments has the other end in $\mathcal{A}$. Then there is no triangle at all, which is a contradiction.