2. (1989 Australian Mathematics Competition) Let the number of distinct divisors of the 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.
2. Because $N(n)$ is odd if and only if $n$ is a perfect square.

Since $44^{2}<1989<45^{2}$,
there are 44 perfect squares among $1,2, \cdots, 1989$, which means there are 44 odd numbers in $N_{(1)}, N_{(2)}, \cdots, N_{(1989)}$, and the rest are even,
thus $N_{(1)}+N_{(2)}+\cdots+N_{(1989)}$ is even.