Bulgarian Mathematical Olympiad

Note that when $x>y+1/2$ run through the interval $(0,n)$ then $x+y$ runs through the interval $(0,2n-1/2)$. Straightforward induction shows that $f$ is bounded in all intervals $(0,2k+1/2)$. When $0<x<y$ it follows that $f$ is also bounded in the intervals $(-2k,0)$. Thus, $f$ is bounded in every bounded subset of $\mathbb{R}$. Let $x\ne 0$. When $y\to 0$ it follows from the boundedness of $f$ and the given equality that $f(x+y)\to f(x)$, i.e. $f$ is continuous at $x$. For $y=-x$ we have $f(x)-f(-x)=f(2x)$. Thus $f(-x)-f(x)=f(-2x)$ and therefore $-f(-2x)=f(2x)=2f(x)$. It follows by induction on $n\ge 3$ that for $x=(n-1)y$ we have $f(ny)=nf(y)$. Hence $f(r)=ar$, where $a=f(1)$ and $r\in {\mathbb{Q}}^{+}$. Since $f$ is odd and continuous we obtain that $f(x)=ax$ for any $x\ne 0$. When $x=y$ we have that $f(0)=0$ implying that $f(x)=ax$ for any $x$. It is easily checked that $f(x)=ax$ is indeed a solution of the problem.