If there are $2 k(k \geqslant 3)$ points on a plane, where no three points are collinear. Draw a line segment between any two points, and color each line segment red or blue. A triangle with three sides of the same color is called a monochromatic triangle, and the number of monochromatic triangles is denoted as $S$. For all possible coloring methods, find the minimum value of $S$.
We will prove that the minimum value of $S$ is $\frac{k(k-1)(k-2)}{3}$.

Since there are $\mathrm{C}_{2 k}^{3}$ triangles, the number of non-monochromatic triangles is $\mathrm{C}_{2 k}^{3}-S$. We call an angle with two adjacent sides (sharing a common vertex) of the same color a monochromatic angle. We will now calculate the number of monochromatic angles.

On one hand, the number of monochromatic angles is $3 S + \left(\mathrm{C}_{2 k}^{3} - S\right) = 2 S + \mathrm{C}_{2 k}^{3}$. On the other hand, for each point $A$, among the $2 k-1$ edges emanating from $A$, if there are $i$ edges of one color, then there are $2 k-1-i$ edges of the other color, where $i=0,1, \cdots, 2 k-1$. The number of monochromatic angles with $A$ as the vertex is
$$
\mathrm{C}_{i}^{2} + \mathrm{C}_{2 k-1-i}^{2} = i^{2} - (2 k-1) i + 2 k^{2} - 3 k + 1,
$$

where $\mathrm{C}_{i}^{2} = 0$ when $i=0,1$.
Let $f(x) = x^{2} - (2 k-1) x + 2 k^{2} - 3 k + 1$, where $x > 0$. Then, when $x \in \left(0, \frac{2 k-1}{2}\right)$, $f(x)$ is strictly decreasing. Therefore, for $i=0,1, \cdots, 2 k-1$, $i^{2} - (2 k-1) i + 2 k^{2} - 3 k + 1$ achieves its minimum value $f(k-1) = (k-1)^{2}$ when $i=k-1$. Thus, the number of monochromatic angles with $A$ as the vertex is at least $(k-1)^{2}$, so we have $2 S + \mathrm{C}_{2 k}^{3} \geqslant 2 k(k-1)^{2}$, which leads to
$$
S \geqslant \frac{1}{2}\left(2 k(k-1)^{2} - \mathrm{C}_{2 k}^{3}\right) = \frac{k(k-1)(k-2)}{3}.
$$

The following example shows that $S$ can achieve the minimum value $\frac{k(k-1)(k-2)}{3}$. Let the $2 k$ points be $A_{1}, A_{2}, \cdots, A_{k}, B_{1}, B_{2}, \cdots, B_{k}$. Color the lines between $A_{1}, A_{2}, \cdots, A_{k}$ red, and color the lines between $B_{1}, B_{2}, \cdots, B_{k}$ red; for all $i, j \in \{1,2, \cdots, k\}$, color the lines between $A_{i}$ and $B_{j}$ blue. Then there are no blue triangles, and the triangles formed by any three points among $A_{1}, A_{2}, \cdots, A_{k}$ are all red, and the triangles formed by any three points among $B_{1}, B_{2}, \cdots, B_{k}$ are all red. Therefore, the number of monochromatic triangles is $2 \mathrm{C}_{k}^{3} = \frac{k(k-1)(k-2)}{3}$.