Romanian Mathematical Olympiad

From $f(1)+f(2)+f(f(1))=4$ follows $f(1)\in \{1,2\}$.

If $f(1)=1$, then $f(2)=2$ and an easy induction shows that $f(n)=n$.

If $f(1)=2$, then $f(2)=1$ and, inductively, $$f(n)=\{\begin{array}{ll}n+1,& \text{if }n\text{ is odd}\\ n-1,& \text{if }n\text{ is even}\end{array}.$$ Both the above found functions fulfill the initial condition.