Let the ten points be $A_{1}, A_{2}, \cdots, A_{10}$. Using these ten points as vertices and the line segments connecting them as edges, we obtain a simple graph $G$ of order 10. Let the degree of point $A_{i} (i=1,2, \cdots, 10)$ be $d_{i}$. Then the total number of edges in graph $G$ is $\frac{1}{2} \sum_{i=1}^{10} d_{i}$.

In graph $G$, we call the figure formed by two line segments sharing a common endpoint an angle. Thus, there are $\sum_{i=1}^{10} \mathrm{C}_{d_{i}}^{2}$ different angles.

By the given condition, for any angle $\angle A_{i} A_{j} A_{k}$ in graph $G$, $A_{i}$ and $A_{k}$ are not adjacent. For any two angles $\angle A_{i_{1}} A_{j_{1}} A_{k_{1}}$ and $\angle A_{i_{2}} A_{j_{2}} A_{k_{2}}$, we have
$\left\{A_{i_{1}}, A_{k_{1}}\right\} \neq\left\{A_{i_{2}}, A_{k_{2}}\right\}$.
Thus, the complement graph of $G$ has at least $\sum_{i=1}^{10} \mathrm{C}_{d_{i}}^{2}$ edges.
Therefore, the total number of edges $S$ satisfies
$$\sum_{i=1}^{10} \mathrm{C}_{d_{i}}^{2}=S \leqslant \mathrm{C}_{10}^{2}-\frac{1}{2} \sum_{i=1}^{10} d_{i} \text {, }$$

which implies $\mathrm{C}_{10}^{2} \geqslant \sum_{i=1}^{10} \mathrm{C}_{d_{i}}^{2}+\frac{1}{2} \sum_{i=1}^{10} d_{i}=\frac{1}{2} \sum_{i=1}^{10} d_{i}^{2}$.
Also, $\frac{1}{2} \sum_{i=1}^{10} d_{i}^{2} \geqslant \frac{1}{2} \times \frac{1}{10}\left(\sum_{i=1}^{10} d_{i}\right)^{2}$, so
$$\mathrm{C}_{10}^{2} \geqslant \frac{1}{2} \times \frac{1}{10}\left(\sum_{i=1}^{10} d_{i}\right)^{2} \Rightarrow \frac{1}{2} \sum_{i=1}^{10} d_{i} \leqslant 15 \text {. }$$

A specific construction is shown in Figure 1.