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 $x \neq 0$. 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 $x \neq 0$. 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.