Let $m \ge n$ be positive integers. Frieder is given $mn$ posters of Linus with different integer dimensions of $k \times l$ with $1 \ge k \ge m$ and $1 \ge l \ge n$. He must put them all up one by one on his bedroom wall without rotating them. Every time he puts up a poster, he can either put it on an empty spot on the wall or on a spot where it entirely covers a single visible poster and does not overlap any other visible poster. Determine the minimal area of the wall that will be covered by posters.
1. **Understanding the Problem:**
   We are given \( mn \) posters with dimensions \( k \times l \) where \( 1 \le k \le m \) and \( 1 \le l \le n \). We need to determine the minimal area of the wall that will be covered by these posters when they are placed without rotation and without overlapping each other.

2. **Reversing the Inequalities:**
   The problem statement contains inequalities \( 1 \ge k \ge m \) and \( 1 \ge l \ge n \), which are incorrect. The correct inequalities should be \( 1 \le k \le m \) and \( 1 \le l \le n \).

3. **Lemma:**
   If \( x < z \) and \( y > w \), then neither of \( (x, y) \) and \( (z, w) \) can fit inside the other. This is because the dimensions do not allow one poster to completely cover the other.

4. **Chains of Posters:**
   We define a chain of posters as a sequence where each poster is placed under the next one. By examining the posters \( (m, 1), (m-1, 2), \ldots, (m-n+1, n) \), we see that each poster cannot fit inside the others due to the lemma. Therefore, there must be at least \( n \) chains.

5. **Pigeonhole Principle:**
   Considering the \( n-1 \) posters \( (m, 2), (m-1, 3), \ldots, (m-n+2, n) \), we use the pigeonhole principle to conclude that at least one of the previous group must be the upper element of a chain. To minimize the area, we notice that the minimal area is achieved by taking the upper elements of the chain as \( (m, 1), (m, 2), \ldots, (m, n) \).

6. **Calculating the Area:**
   The total area covered by these \( n \) chains is:
   \[
   A = m \cdot 1 + m \cdot 2 + \cdots + m \cdot n = m \sum_{k=1}^{n} k = m \cdot \frac{n(n+1)}{2}
   \]

Conclusion:
The minimal area of the wall that will be covered by the posters is:
\[
\boxed{m \cdot \frac{n(n+1)}{2}}
\]