$1 \cdot 71$ 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 $N(1)+N(2)+\cdots+$ $N(1989)$ is odd or even.
[Solution] Since $d$ is a divisor of $n$, $\frac{n}{d}$ is also a divisor of $n$, and only when $n$ is a perfect square, i.e., $d=\sqrt{n}$, does $d=\frac{n}{d}$. Therefore, in general, the divisors of $n$ appear in pairs ($d$ and $\frac{n}{d}$ form a pair), and only when $n$ is a perfect square, there is one divisor $\sqrt{n}$ that cannot be paired. This indicates that $N(n)$ is odd if and only if $n$ is a perfect square.

Since $45^{2}>1989>44^{2}$, there are 44 perfect squares in the sequence $1,2, \cdots, 1989$, meaning there are 44 odd numbers in the sum, thus the sum is even.