Let $n$ be a positive integer. A square $ABCD$ is partitioned into $n^2$ unit squares. Each of them is divided into two triangles by the diagonal parallel to $BD$. Some of the vertices of the unit squares are colored red in such a way that each of these $2n^2$ triangles contains at least one red vertex. Find the least number of red vertices.

[i](4th Middle European Mathematical Olympiad, Team Competition, Problem 4)[/i]
To solve this problem, we need to ensure that each of the \(2n^2\) triangles formed by the diagonals of the unit squares contains at least one red vertex. We aim to find the minimum number of red vertices required to achieve this.

1. **Understanding the Problem:**
   - A square \(ABCD\) is divided into \(n^2\) unit squares.
   - Each unit square is further divided into two triangles by the diagonal parallel to \(BD\).
   - We need to color some vertices red such that each of these \(2n^2\) triangles contains at least one red vertex.

2. **Analyzing the Triangles:**
   - Each unit square has 4 vertices.
   - The diagonal divides each unit square into two triangles.
   - Each triangle shares vertices with adjacent triangles.

3. **Optimal Coloring Strategy:**
   - To minimize the number of red vertices, we should try to color vertices that are shared by multiple triangles.
   - Consider the vertices of the unit squares in a grid pattern. We can color vertices in a checkerboard pattern, where vertices at positions \((i, j)\) with \(i + j\) even are colored red.

4. **Checkerboard Pattern:**
   - In a checkerboard pattern, each vertex at \((i, j)\) with \(i + j\) even is colored red.
   - This ensures that each triangle has at least one red vertex because each triangle will have at least one vertex with \(i + j\) even.

5. **Counting the Red Vertices:**
   - In an \(n \times n\) grid, there are \(n^2\) vertices.
   - Half of these vertices will have \(i + j\) even, and the other half will have \(i + j\) odd.
   - Therefore, the number of red vertices is \(\left\lceil \frac{n^2}{2} \right\rceil\).

6. **Verification:**
   - For \(n\) even, \(\frac{n^2}{2}\) is an integer.
   - For \(n\) odd, \(\frac{n^2}{2}\) is not an integer, so we take the ceiling function to ensure all triangles are covered.

Thus, the least number of red vertices required is \(\left\lceil \frac{n^2}{2} \right\rceil\).

The final answer is \(\boxed{\left\lceil \frac{n^2}{2} \right\rceil}\).