We prove that $f(x)=a x+b$ where $a \in \mathbb{Q}$ and $b \in \mathbb{R}$. These functions obviously satisfy the conditions. Suppose that a function $f(x)$ fulfills the required properties. For an arbitrary rational $q$, consider the function $g_{q}(x)=f(x+q)-f(x)$. This is a continuous function which attains only rational values, therefore $g_{q}$ is constant. Set $a=f(1)-f(0)$ and $b=f(0)$. Let $n$ be an arbitrary positive integer and let $r=f(1 / n)-f(0)$. Since $f(x+1 / n)-f(x)=f(1 / n)-f(0)=r$ for all $x$, we have $$f(k / n)-f(0)=(f(1 / n)-f(0))+(f(2 / n)-f(1 / n))+\ldots+(f(k / n)-f((k-1) / n)=k r$$ and $f(-k / n)-f(0)=-(f(0)-f(-1 / n))-(f(-1 / n)-f(-2 / n))-\ldots-(f(-(k-1) / n)-f(-k / n)=-k r$ for $k \geq 1$. In the case $k=n$ we get $a=f(1)-f(0)=n r$, so $r=a / n$. Hence, $f(k / n)-f(0)=k r=a k / n$ and then $f(k / n)=a \cdot k / n+b$ for all integers $k$ and $n>0$. So, we have $f(x)=a x+b$ for all rational $x$. Since the function $f$ is continuous and the rational numbers form a dense subset of $\mathbb{R}$, the same holds for all real $x$.