Let $x \in \mathbb{Q}^+$, $x = \frac{a}{b}$, $(a, b) = 1$, $a > 0$, $b > 0$. Now let us apply Euclid's algorithm for $a$ and $b$. Here $r_0 = a$, $r_1 = b$. $r_{j-1} = q_j r_j + r_{j+1}$, $j = 1, 2, \dots, n$. There exists $n = n(x)$, such that $r_n \neq 0$ and $r_{n+1} = 0$.

Consider the function $f: \mathbb{Q} \to \{-1; 1\}$ defined by
$$f(x) = \begin{cases} (-1)^{n(x)} & x > 0 \\ 1 & x = 0 \\ (-1)^{n(-x)+1} & x < 0 \end{cases}$$

Now let us prove that $f(x)$ satisfies the given condition.

I. Indeed let $x + y = 0$, $x \neq y$ and $x > 0, y < 0$. $f(x) = (-1)^{n(x)}$, $f(y) = f(-x) = (-1)^{n(x)+1} = -f(x)$. Then from this $f(x) \cdot f(y) = -1$, which means $x$ and $y$ have different colors.

II. Let $x + y = 1$, $x \neq y$. $x, y \in \mathbb{Q}$. Then at least one of $x, y$ is positive. Let $x > 0$, $x = \frac{a}{b}$. Then $y = \frac{b-a}{b}$. If $y < 0$, then $f(y) = (-1)^{n(-y)+1} = (-1)^{n(\frac{a-b}{b})+1}$. Considering $n(\frac{a}{b}) = n(\frac{a-b}{b})$, we have $f(x) \cdot f(y) = -1$. If $y > 0$, then $0 < x < \frac{1}{2} < y < 1$. From Euclid's algorithm $n(y) = n(x) + 1$. $f(x) \cdot f(y) = -1$.

III. Let $xy = 1$, then $x, y$ have same sign. In this case $n(y) = n(x) + 1$, that implies $f(x)f(y) = -1$.