Solution:
We need $n^{2}-\frac{1989}{n}$ to be a perfect square, so $n \mid 1989$. Also, this perfect square would be less than $n^{2}$, so it would be at most $(n-1)^{2}=n^{2}-2 n+1$. Thus,

$$\frac{1989}{n} \geq 2 n-1 \Longrightarrow 1989 \geq 2 n^{2}-n$$

so $n \leq 31$. Moreover, we need
$$n^{2} \geq \frac{1989}{n} \Longrightarrow n^{3} \geq 1989$$
so $n \geq 13$. Factoring gives $1989=3^{2} \cdot 13 \cdot 17$, which means the only possible values of $n$ are 13 and 17. Checking both gives that only $n=13$ works. (In fact, $\frac{13^{3}-1989}{13}=4^{2}$.)

Solution:
If $\frac{n^{3}-1989}{n}=d^{2}$ then $n^{3}-n d^{2}=1989$. Factorizing gives

$$(n-d) n(n+d)=3^{2} \times 13 \times 17$$

We can easily see that $n=13, d=4$ works since $1989=9 \times 13 \times 17$.