Solution:

Consider all $a, b$ in the range $1, 2, 3, \ldots, N^{2}$. There are $N^{4}$ possible pairs of values. But $[a^{3/2}]$ and $[b^{3/2}]$ are in the range $1, 2, \ldots, N^{3}$, so their sum is in the range $1, 2, \ldots, 2N^{3}$. Hence one of these values has at least $N/2$ solutions. By taking $N$ sufficiently large we can get $a_{1} > 1980$ solutions for some $N_{1} \leq 2N^{3}$. But now by taking $N$ sufficiently large we can get $a_{2} > a_{1}$ solutions for some $N_{2}$. Since $a_{2} \neq a_{1}$, we must have $N_{2} \neq N_{1}$. In other words, we have a different $n$, also with $>1980$ solutions. Continuing, we get an infinite sequence of distinct $n$ each with at least $1980$ solutions.