24.8. (Jury, CSSR, 79). Let $M$ be a subset of the set of all pairs of natural numbers $i<k$, not exceeding a given natural number $n \geqslant 2$. If a pair $i<k$ belongs to the set $M$, then no pair $k<m$ belongs to it. What is the maximum number of pairs that can be in the set $M$?
24.8. Let $A$ and $B$ denote the sets of all smaller and, respectively, larger numbers that form the pairs of the set $M$. Then, by the condition of the problem, no element from $B$ is contained in the set $A$, and therefore, $A \cap B=\varnothing$. Let the number of elements in the sets $A$ and $B$ be $a$ and $b$ respectively. Then $a+b=n$. The number of pairs in $M$ is equal to the number of ways to choose one element from $A$ and one element from $B$, which is $a \cdot b$. To maximize the number of pairs, we need to maximize the product $a \cdot b$ under the condition $a + b = n$.

If the number $n$ is even, then the maximum product is achieved when $a = b = n / 2$:

$$
M=\{(i ; j) \mid i \leq n / 2, j > n / 2\}
$$

(then $M$ consists of $n^{2} / 4$ elements). If the number $n$ is odd, then it is sufficient to take

$$
M=\{(i ; j) \mid i \leq (n-1) / 2, j > (n-1) / 2\}
$$

(then $M$ consists of

$$
\frac{n-1}{2} \cdot \frac{n+1}{2}=\frac{n^{2}}{4}-\frac{1}{4}=\left[\frac{n^{2}}{4}\right]
$$

elements). Thus, in both the case of even and odd $n$, the maximum possible number of elements in the set $M$ is $\left[n^{2} / 4\right]$.