Answer: 720.

Solution. Let's inscribe the given polygon $K_{1} K_{2} \ldots K_{20}$ in a circle. Each trapezoid is defined by a pair of parallel chords with endpoints at points $K_{1}, \ldots, K_{20}$.

Consider a chord connecting two adjacent vertices of the polygon, for example, $K_{6} K_{7}$. There are 9 more chords parallel to it ( $K_{5} K_{8}$ and so on), i.e., we get a set of 10 chords parallel to each other. From these, we can form $C_{10}^{2}=45$ pairs of parallel segments.

We will get similar sets if we consider all chords parallel to $K_{1} K_{2}, \ldots, K_{10} K_{11}$, a total of 10 such sets.

Now, let's take a chord connecting vertices that are one apart from each other, for example, $K_{6} K_{8}$. There are 8 more chords parallel to it ( $K_{5} K_{9}$ and so on), i.e., we get a set of 9 chords parallel to each other. From these, we can form $C_{9}^{2}=36$ pairs of parallel segments. There are also 10 such sets.

In total, we get $10 \cdot 45 + 10 \cdot 36 = 810$ quadrilaterals. With this method of counting, rectangles have been counted twice. Note that both diagonals of an inscribed rectangle are diameters, and there are 10 diameters with vertices at the given points, thus we have $C_{10}^{2}=45$ rectangles. Subtracting twice the number of rectangles, we get $810 - 90 = 720$ quadrilaterals that have at least one pair of parallel sides.
5. Given a regular 20-gon $M$. Find the number of quadruples of vertices of this 20-gon that are the vertices of trapezoids.