$4 \cdot 2$ Let the number of distinct divisors of a positive integer $n$ be $N(n)$, for example, 24 has divisors 1, $2,3,4,6,8,12,24$, so $N_{(24)}=8$. Determine whether the sum
$$
N_{(1)}+N_{(2)}+\cdots+N_{(1989)}
$$

is odd or even.
(Australian Mathematics Competition, 1989)
［Solution］Because $N(n)$ is odd if and only if $n$ is a perfect square.
Also, since
$$
44^{2}<1989<45^{2} \text {, }
$$

there are 44 perfect squares among the numbers $1,2, \cdots, 1989$, which means that among the numbers
$$
N_{(1)}, N_{(2)}, \cdots, N_{(1989)}
$$

there are 44 odd numbers, and the rest are even,
thus $N_{(1)}+N_{(2)}+\cdots+N_{(1989)}$ is even.