Brazilian Mathematical Olympiad

First of all, $f$ is injective: indeed, $f(x) = f(y) \implies f(2f(x)) = f(2f(y)) \iff x + 1998 = y + 1998 \iff x = y$.

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.