For each positive integer $m$, we define $L_m$ as the figure that is obtained by overlapping two $1 \times m$ and $m \times 1$ rectangles in such a way that they coincide at the $1 \times 1$ square at their ends, as shown in the figure.

[asy]
pair h = (1, 0), v = (0, 1), o = (0, 0);
for(int i = 1; i < 5; ++i)
{
  o = (i*i/2 + i, 0);
  draw(o -- o + i*v -- o + i*v + h -- o + h + v -- o + i*h + v -- o + i*h -- cycle);
  string s = "$L_" + (string)(i) + "$";
  label(s, o + ((i / 2), -1));
  for(int j = 1; j < i; ++j)
  {
    draw(o + j*v -- o + j*v + h);
    draw(o + j*h -- o + j*h + v);
  }
}
label("...", (18, 0.5));
[/asy]

Using some figures $L_{m_1}, L_{m_2}, \dots, L_{m_k}$, we cover an $n \times n$ board completely, in such a way that the edges of the figure coincide with lines in the board. Among all possible coverings of the board, find the minimal possible value of $m_1 + m_2 + \dots + m_k$.

Note: In covering the board, the figures may be rotated or reflected, and they may overlap or not be completely contained within the board.
1. **Understanding the Problem:**
   We need to cover an \( n \times n \) board using figures \( L_{m_1}, L_{m_2}, \dots, L_{m_k} \). Each \( L_m \) is formed by overlapping a \( 1 \times m \) and an \( m \times 1 \) rectangle at a \( 1 \times 1 \) square. The goal is to find the minimal possible value of \( m_1 + m_2 + \dots + m_k \).

2. **Initial Observation:**
   Each \( L_m \) covers \( 2m - 1 \) cells on the board. This is because the \( 1 \times m \) and \( m \times 1 \) rectangles overlap at one cell, so the total number of unique cells covered is \( m + m - 1 = 2m - 1 \).

3. **Counting the Total Coverage:**
   To cover the entire \( n \times n \) board, we need to ensure that the sum of the areas covered by the \( L \)-shaped figures is at least \( n^2 \). Therefore, we need:
   \[
   (2m_1 - 1) + (2m_2 - 1) + \dots + (2m_k - 1) \geq n^2
   \]

4. **Simplifying the Inequality:**
   Simplifying the left-hand side, we get:
   \[
   2(m_1 + m_2 + \dots + m_k) - k \geq n^2
   \]
   Rearranging this inequality, we obtain:
   \[
   2(m_1 + m_2 + \dots + m_k) \geq n^2 + k
   \]

5. **Minimizing \( m_1 + m_2 + \dots + m_k \):**
   To minimize \( m_1 + m_2 + \dots + m_k \), we need to consider the smallest possible value of \( k \). The minimum number of \( L \)-shaped figures required to cover the board is \( n \), because each \( L \)-shaped figure can cover at most \( 2m - 1 \) cells, and we need to cover \( n^2 \) cells.

6. **Substituting \( k = n \):**
   Substituting \( k = n \) into the inequality, we get:
   \[
   2(m_1 + m_2 + \dots + m_k) \geq n^2 + n
   \]
   Dividing both sides by 2, we obtain:
   \[
   m_1 + m_2 + \dots + m_k \geq \frac{n^2 + n}{2}
   \]

7. **Achieving the Minimum Sum:**
   The sum \( 1 + 2 + \dots + n \) is given by the formula for the sum of the first \( n \) positive integers:
   \[
   1 + 2 + \dots + n = \frac{n(n+1)}{2}
   \]
   This matches the lower bound we derived, showing that it is possible to achieve this sum.

8. **Conclusion:**
   Therefore, the minimal possible value of \( m_1 + m_2 + \dots + m_k \) is \( \frac{n(n+1)}{2} \).

The final answer is \( \boxed{\frac{n(n+1)}{2}} \)