Let $n$ be a positive integer. A regular hexagon with side length $n$ is divided into equilateral triangles with side length $1$ by lines parallel to its sides. 
Find the number of regular hexagons all of whose vertices are among the vertices of those equilateral triangles.

[i]UK - Sahl Khan[/i]
1. **Understanding the Problem:**
   We are given a regular hexagon with side length \( n \) that is divided into smaller equilateral triangles with side length 1. We need to find the number of regular hexagons, all of whose vertices are among the vertices of these equilateral triangles.

2. **Hexagon Center and Vertices:**
   Consider an arbitrary hexagon with center \( P \) and vertices \( A, B, C, D, E, F \). Since \( A, B, C, D, E, F \) are vertices of the equilateral triangles, \( P \) must also be a vertex of these triangles. Therefore, we can count hexagons by their center vertex \( P \).

3. **Labeling Vertices:**
   Label each vertex \( P \) of a triangle with the number \( k \), where \( k \) is the minimum number of equilateral triangle edges one must travel to get from any edge vertex to \( P \). Vertices on the edge are labeled with 0, vertices one step inside are labeled with 1, and so on.

4. **Counting Hexagons Centered at \( P \):**
   A hexagon centered at \( P \) either has exactly one vertex contained in the interior of the drawn angle or two vertices on the angle's boundaries. If \( P \)'s label is \( k \), the number of points in the interior is \( \binom{k}{2} \), and the number of points on one boundary is \( k \). So there are \( \binom{k+1}{2} \) hexagons centered at \( P \).

5. **Summing Up Hexagons:**
   We need to sum up the value of \( \binom{k+1}{2} \) for each vertex labeled with \( k \). The center of the hexagon is labeled \( n \), and the points labeled \( n-k \) are those lying on the edges of a regular hexagon of side length \( k \) concentric with the original hexagon. There are \( 6k \) such points.

6. **Total Number of Hexagons:**
   The total number of hexagons is given by:
   \[
   \sum_{k=0}^{n} 6k \binom{k+1}{2}
   \]
   Simplifying this, we get:
   \[
   \sum_{k=0}^{n} 6k \frac{(k+1)k}{2} = 3 \sum_{k=0}^{n} k^2 (k+1)
   \]
   This sum can be evaluated as:
   \[
   3 \sum_{k=0}^{n} (k^3 + k^2) = 3 \left( \sum_{k=0}^{n} k^3 + \sum_{k=0}^{n} k^2 \right)
   \]
   Using the formulas for the sum of cubes and squares:
   \[
   \sum_{k=0}^{n} k^3 = \left( \frac{n(n+1)}{2} \right)^2 \quad \text{and} \quad \sum_{k=0}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}
   \]
   Substituting these in, we get:
   \[
   3 \left( \left( \frac{n(n+1)}{2} \right)^2 + \frac{n(n+1)(2n+1)}{6} \right)
   \]
   Simplifying further, we get:
   \[
   3 \left( \frac{n^2(n+1)^2}{4} + \frac{n(n+1)(2n+1)}{6} \right)
   \]
   Combining the terms, we get:
   \[
   \frac{3n^2(n+1)^2}{4} + \frac{n(n+1)(2n+1)}{2}
   \]
   Simplifying this, we get:
   \[
   \frac{(n^2 + n)^2}{4}
   \]

The final answer is \(\boxed{\frac{(n^2 + n)^2}{4}}\).