4. (1) From $(a, b)=2 \times 3^{3} \times 37$, we know $37 \mid[a, b]$, which means $37 \mid n$, so $n \geqslant 37$. When $n \geqslant 37$, $1998 \mid n!$, at this time, taking $a=1998, b=n!$ will suffice. Therefore, the positive integers $n$ that satisfy the condition are all positive integers not less than 37.
(2) Let $a=1998 x, b=1998 y$, then $(x, y)=1$. When $n \geqslant 37$, we have $x y=2^{33} \times 3^{14} \times 5^{8} \times 7^{5} \times 11^{3} \times 13^{2} \times 17^{2} \times 19 \times 23 \times 29 \times 31 \times\left(\frac{n!}{37!}\right)$.

Thus, when $37 \leqslant n \leqslant 40$, the number of pairs $(x, y)$ is $\frac{1}{2} \times 2^{11}=1024$ pairs, and when $n \geqslant 41$, there are at least $\frac{1}{2} \times 2^{12}=2048$ pairs, hence $37 \leqslant n \leqslant 40$.