XX OBM

First of all, $f$ is injective: indeed, $f(x)=f(y)⟹f(2f(x))=f(2f(y))⟺x+1998=y+1998⟺x=y$.

Moreover, $f$ takes all integer values bigger than $1997$: if $m\ge 1998$ then $m=1998+k$ for some $k\ge 0$ and $f(2f(k))=k+1998=m$.

Notice that the values of $m$ such that $f(m)\ge 1998$ are all even; since $f$ is injective, $f(x)$ can't be bigger than $1997$ for every $x$ odd. This means that all odd numbers are taken by $f$ to numbers not exceeding $1998$. By the pigeon hole principle, there are two different odd numbers $x$ and $y$ such that $f(x)=f(y)$, contradiction. So there is no such function.