Let $n$ be a positive integer. A regular hexagon with side length $n$ is divided into a series of equilateral triangles with side length 1, whose sides are parallel to those of the hexagon. Find the number of regular hexagons whose vertices are the vertices of these equilateral triangles.
We call a regular hexagon a lattice hexagon if its vertices are the vertices of a triangle and its six sides are parallel to the sides of a larger hexagon. For any regular hexagon $H$, we can always find a lattice hexagon $G$ such that $H$ is inscribed in $G$, meaning the six vertices of $H$ lie exactly on the six sides of $G$, as shown in Figure 5. Conversely, for a lattice regular hexagon $H$ with side length $m$, there are exactly $m$ inscribed regular hexagons $G$.

Next, we calculate the number of lattice regular hexagons with side length $m$. Since the center of such a lattice regular hexagon must be no more than $n-m$ units away from the center of the larger hexagon (otherwise, the vertices of the hexagon would be outside the larger hexagon), we only need to count the lattice points within a regular hexagon centered at the center of the larger hexagon with side length $n-m$, including the boundary, which is $6 \sum_{i=1}^{n-m} i+1=3(n-m)(n-m+1)+1$.

Therefore, the number of hexagons that meet the criteria is
$$
\begin{array}{l}
\quad N=\sum_{m=1}^{n}(3(n-m)(n-m+1)+1) m \\
=\left(3 n^{2}+3 n\right) \sum_{m=1}^{n} m-3(2 n+1) \sum_{m=1}^{n} m^{2}+3 \sum_{m=1}^{n} m^{3} \\
=\left(\frac{n(n+1)}{2}\right)^{2}
\end{array}
$$