1. Let $n=\underbrace{111 \cdots 11}_{1999 \uparrow 1}, f(n)=90 n^{2000}+20 n+$ 1997. Then the remainder when $f(n)$ is divided by 3 is
Ni.1.1.
A natural number $a$ has the same remainder when divided by 3 as its digit sum $S(a)$ when divided by 3. Therefore, $n=\underbrace{11 \cdots 111}_{1999 \uparrow 1}$ has the same remainder when divided by 3 as 1999, which is 1.
$90 n^{2000}$ has a remainder of 0 when divided by 3, and $20 n$ has a remainder of 2 when divided by 3.

Thus, the remainder of $f(n)=90 n^{2000}+20 n+1997$ when divided by 3 is the same as the remainder of $2+1997=1999$ when divided by 3, which is 1.