Zlobin C.A.

The greatest common divisor of natural numbers $m$ and $n$ is 1. What is the greatest possible value of $\gcd(m+2000 n, n+2000 m) ?$
Let $a=2000 m+n, b=2000 n+m, d=$ GCD $(a, b)$. Then $d$ divides the numbers $2000 a-b=\left(2000^{2}-1\right) m$ and $2000 b-a=\left(2000^{2}-1\right) n$. Since $m$ and $n$ are coprime, $d$ divides $2000^{2}-1$.

On the other hand, for $m=2000^{2}-2000-1, n=1$, we get $a=\left(2000^{2}-1\right)(2000-1), b=2000^{2}-1=d$.

## Answer

3999999.