We''ll show that $f(x)=c$ for $x \in N$. Denote $f(1)$ by $c$. First of all: by taking $x=1,y=2$ we obtain $f(2)=f(1)=c$ Now let's plug in $x=y=3$,in this case $f(3)=f(2)=c$. Let's prove by induction that for every $x>1$ $f(x)=c$ We have $f(1)=f(2)=f(3)=c$. Assume that for every $1 \leq x \leq n$ (where $n \geq 3$) we have $f(n)=c$. Let us consider $f(n+1)$. Let $t$ be such number in $<1;n>$ that $(n+1)+t$ is a multiple of $3$. Then \[f(\frac{n+1+t}{3})=\frac{f(n+1)+f(t)}{2}\] but as $t \leq n$ and $\frac{n+1+t}{3}\leq n$ we get $f(n+1)=c$ which means that \[f(x)=c\] for every $x \geq 1$. Now let $k$ be a negative number. There exists $l>0$ such that $l+k>0$ and $l+k$ is a multiple of $3$. Puting $x=l,y=k$ in our equation leads us to $f(k)=c$. Finally we got $f(x)=c$ for a constant $c \in Q$.