Let $c$ denote the constant term in $Q(x)$. Suppose on the contrary that there exists $n$ such that $Q(P(n))=1$. Clearly we have $P(x)=(x-a)(x-a-1997)R(x)$ for some polynomial $R(x)$. Regardless of the parity of $a$ exactly one of $x-a$ and $x-a-1997$ is even, so that $2|P(x)$. Then $2|P(n)$ so $2|Q(P(n))-c=1-c$ and so $c$ is odd. Hence $Q(2x)$ is odd since we have all terms even apart from the odd constant term $c$. But this contradicts $Q(1998)=2000$. Hence no $n$ exists.