Example 1. (4th Ling Ling Camp Test Question) Given 1989 points, no three of which are collinear. Divide them into 30 groups with different numbers of points. From any three different groups, take one point from each to form a triangle. To maximize the total number of such triangles, how many points should each group have?
Let the number of points in each group be $x_{\mathrm{K}} (k=1,2, \cdots, 30)$, all $x_{\mathbf{k}}$ are distinct, $\sum_{\mathrm{k}=1}^{30} x_{\mathrm{k}}=1989$, the total number of triangles is $F\left(x_{1}, x_{2}, \cdots, x_{2}, 0\right)$
$$
=\sum_{1 \leqslant i<j<k \leqslant 30} x_{1} x_{1} x_{k},
$$

Obviously, this problem is about finding the maximum value of the symmetric function $F$ for the type $N(1989, 1)$.
Note: $F\left(a, b, c_{3}, \cdots, c_{30}\right)$
$$
\begin{array}{l}
=a b \sum_{3 \leqslant k \leqslant 30} c z \\
+(a+b) \sum_{3 \leqslant j<k \leqslant 30} c_{1} c_{k} \\
+\sum_{3 \leqslant i<j<k \leqslant 30}^{c 1 c, c 2}
\end{array}
$$

Since $a \leqslant b$, it is clear that $(a-1)(b+1)=a b-(b+1-a) < a b$, and it is easy to see that $F\left(a-1, b+1, c_{3}, \cdots, c_{30}\right) < F\left(a, b, c_{3}, \cdots, c_{30}\right)$.
By Theorem 1 (here $S=1989, d=1, n=30$. From $S-\frac{1}{2} n(n-1) d=1554$ divided by 30, we get $q=51$, $r=24$), we know that the number of points in each group should be:
$51,52, \cdots, 56,58,59, \ldots, 81$.