Let $n \geq 2$ be an integer. Find the maximal cardinality of a set $M$ of pairs $(j, k)$ of integers, $1 \leq j < k \leq n$, with the following property: If $(j, k) \in M$, then $(k,m) \not \in M$ for any $m.$
1. **Graph Representation and Triangle-Free Property**:
   - Consider the graph \( G = (V, M) \) where \( V = \{1, 2, \ldots, n\} \) and \( M \) is the set of edges (pairs \((j, k)\)).
   - The problem states that if \((j, k) \in M\), then \((k, m) \not\in M\) for any \( m \). This implies that \( G \) contains no directed path of length 2, and hence no triangle.
   - By Turán's theorem, the maximum number of edges in a triangle-free graph on \( n \) vertices is given by:
     \[
     |M| \leq \left\lfloor \frac{n^2}{4} \right\rfloor
     \]
   - This bound is achieved by a complete bipartite graph \( K_{\lfloor n/2 \rfloor, \lceil n/2 \rceil} \), where the vertex set \( V \) is partitioned into two disjoint sets \( A \) and \( B \) such that \( A = \{1, \ldots, \lfloor n/2 \rfloor\} \) and \( B = \{\lceil n/2 \rceil, \ldots, n\} \).

2. **Directed Graph and Bipartite Structure**:
   - Consider the directed graph \( \vec{G} = (V, M) \) where \( M \) is the set of directed edges \((j, k)\).
   - Since there is no directed path of length 2, the graph can be made bipartite by removing isolated vertices.
   - Let \( A = \pi_1(M) \) and \( B = \pi_2(M) \), where \(\pi_1(j, k) = j\) and \(\pi_2(j, k) = k\). This means \( A \) contains the starting vertices of the edges and \( B \) contains the ending vertices.
   - Denote \( a = |A| \) and \( b = |B| \). Since \( A \cap B = \emptyset \) and \( A \cup B \subseteq V \), we have \( a + b \leq n \).
   - The number of edges \( |M| \) is maximized when \( |M| = ab \). The product \( ab \) is maximized when \( a \) and \( b \) are as close as possible to \( n/2 \), i.e., \( a = \lfloor n/2 \rfloor \) and \( b = \lceil n/2 \rceil \).

3. **Conclusion**:
   - Therefore, the maximal cardinality of the set \( M \) is:
     \[
     |M| = \left\lfloor \frac{n^2}{4} \right\rfloor
     \]

The final answer is \( \boxed{ \left\lfloor \frac{n^2}{4} \right\rfloor } \)