19-th Macedonian Mathematical Olympiad

Let $f(0)=a$, then $f(a)=f(f(0))=2$, $f(2)=f(f(a))=a+2$. Continuing this procedure we get that $f(2k)=a+2k$ and $f(a+2k)=2k+2$. We get $2k+2=f(a+2k)<f(a)+f(2k)=2+a+2k$, from where we get that $a>0$. If we put $x=y=a$ we get $a+2a=f(2a)<f(a)+f(a)=4$ so $3a<4$. i.e. $a=1$. Hence using $f(2k)=a+2k$ and $f(a+2k)=2k+2$ we get $f(x)=x+1$ for all natural numbers $x$.

For $x=y=\frac{1}{2}$ in the inequality we get $2=f(1)=f(\frac{1}{2}+\frac{1}{2})<2f(\frac{1}{2})$, so $f(\frac{1}{2})>1$. On the other hand $1+f(\frac{1}{2})=f(f(\frac{1}{2}))=2$, so $f(\frac{1}{2})=1$, which is a contradiction. It follows that there exists no function satisfying the required conditions.