$7 \cdot 60$ Given a natural number $n \geqslant 2$. Let $M$ be a subset of the set $\{(i, k) \mid i, k \in N, i < k \leqslant n\}$. It is known that if the pair $(i, k), i < k$, belongs to the set $M$, then any pair $(k, m), k < m$, does not belong to $M$. Find the maximum number of elements in the set $M$.
［Solution］Let
$$
A=\{i \mid(i, k) \in M\}, B=\{k \mid(i, k) \in M\} .
$$

According to the given information, any element in $B$ is not in $A$, i.e., $A \cap B=\varnothing$. Let $|A|=a$, $|B|=b$, then we have $a+b \leqslant n$, and the smaller number in the pairs of set $M$ has $a$ different choices, while the larger number has $b$ different choices. Therefore, the number of elements in $M$ is at most
$$
a b \leqslant a(n-a) \leqslant\left(\frac{a+n-a}{2}\right)^{2}=\frac{n^{2}}{4} .
$$

Since $a b \in \mathbb{N}$, we have $a b \leqslant\left[\frac{n^{2}}{4}\right]$.
On the other hand, when $n$ is even, let
$$
M_{0}=\left\{(i, j) \left\lvert\, i \leqslant \frac{n}{2}<j \leqslant n\right.\right\},
$$

then $\left|M_{0}\right|=\frac{n^{2}}{4}$ and $M_{0}$ satisfies the requirements of the problem; when $n$ is odd, let
$$
M_{0}=\left\{(i, j) \left\lvert\, i<\frac{n}{2}<j \leqslant n\right.\right\},
$$

then $\left|M_{0}\right|=\left[\frac{n^{2}}{4}\right]$ and $M_{0}$ satisfies the requirements of the problem.
In summary, the maximum number of elements in $M$ is $\left[\frac{n^{2}}{4}\right]$.