jmc

Let $m=2^{1998}-1$ and $n=2^{1989}-1$. Then, $2^{9}n=2^9(2^{1989}-1)=2^{1998}-2^9=m-(2^9-1)$. By the Euclidean algorithm, it follows that $$\begin{array}{rl}\text{gcd}(m,n)& =\text{gcd}(n,m-2^{9}n)\\ & =\text{gcd}(n,2^9-1).\end{array}$$Since $9$ divides $1998$, by the difference of odd powers factorization, it follows that $2^{1989}-1$ is divisible by $2^9-1$. Thus, the greatest common divisor of $m$ and $n$ is $2^9-1=\boxed{511}$.