Problem 1122

Notice that, $\sum_{i=1}^{k} m_{i}=\mathrm{C}_{n}^{2}$.
Let $\Delta(S)$ denote the number of isosceles triangles (including degenerate cases of two points and their midpoint) formed by triples of points in the set $S$, where each equilateral triangle is counted three times. For $D \in S$, let $m_{i}(D)$ denote the number of points in the set $S$ that are at a distance $d_{i}(i=1,2, \cdots, k)$ from $D$.

Since point $D$ lies on the perpendicular bisector of $P Q$ if and only if $|D P|=|D Q|$, we have
$$\Delta(S)=\sum_{D \in S} \sum_{i=1}^{k} \mathrm{C}_{m_{i}(D)}^{2}=\sum_{i=1}^{k} \sum_{D \in S} \mathrm{C}_{m_{i}(D)}^{2} .$$

Notice that, $\sum_{D \in S} m_{i}(D)=2 m_{i}$.
Applying the Cauchy-Schwarz inequality to the right-hand side of (1), we get
$$\begin{array}{l} \Delta(S)=\sum_{i=1}^{k} \sum_{D \in S} \frac{m_{i}^{2}(D)-m_{i}(D)}{2} \\ \geqslant \frac{2}{n} \sum_{i=1}^{k} m_{i}^{2}-\sum_{i=1}^{k} m_{i}=\frac{2}{n} \sum_{i=1}^{k} m_{i}^{2}-\mathrm{C}_{n}^{2} . \\ \text { Hence } \sum_{i=1}^{k} m_{i}^{2} \leqslant \frac{n}{2}\left(\Delta(S)+\mathrm{C}_{n}^{2}\right) . \end{array}$$

Furthermore, each line segment $P Q$ can be the base of at most three isosceles triangles determined by the set $S$, otherwise, the perpendicular bisector of $P Q$ would pass through at least four points in $S$, which is a contradiction.
Therefore, $\Delta(S) \leqslant 3 \mathrm{C}_{n}^{2}$.
From (2) and (3), we get $\sum_{i=1}^{k} m_{i}^{2} \leqslant n^{3}-n^{2}$.