Romanian Mathematical Olympiad

If $g(y_1) = g(y_2)$, the first equality yields $y_1 = y_2$, hence $g$ is injective. Analogously, $f$ is injective, as well.
Plugging $y = 0$ in the first equality gives $f(g(x) + g(0)) = f(g(x))$, hence $g(x) + g(0) = g(x)$, so that $g(0) = 0$; similarly, $f(0) = 0$.
Plugging $x = 0$ in both equalities yields $f(g(y)) = g(f(y)) = y$, for all $y$, therefore $f$ and $g$ are bijective and $g = f^{-1}$.
The initial equalities become
$$f(g(x) + g(y)) = x + y, \quad g(f(x) + f(y)) = x + y,$$
for all $x, y \in \mathbb{Q}$, and we deduce
$$g(x + y) = g(x) + g(y), \quad f(x + y) = f(x) + f(y),$$
for all $x, y \in \mathbb{Q}$. Setting $a = f(1), b = g(1)$, it is easy to prove that $f(x) = ax, g(x) = bx$, for all $x \in \mathbb{Q}$.
Since $g = f^{-1}$, we obtain $ab = 1$. Finally, the functions are
$$f(x) = ax, \quad g(x) = \frac{x}{a},$$
where $a \in \mathbb{Q}^*$, and it is easy to see that they satisfy the initial equalities.