By replacing $x \leftarrow x + \tfrac{1}{4} $, $y \leftarrow y + \tfrac{1}{4} $ we can write the given as $$f(x)+f(y)=f(x + \lfloor y + \tfrac{1}{2} \rfloor ).$$Putting $x=y=0$ yields $f(0)=0$. Next, plugging in $x=0$ gives $$f(y) = f(\lfloor y + \tfrac{1}{2} \rfloor ).$$Hence $f$ is uniquely determined by its values on the integers. Now if we plug in $x,y \in \mathbb{Z}$ we get $f(x)+f(y)=f(x+y)$, hence $f$ is additive on the integers. Thus $f(n)=cn$ for $n \in \mathbb{Z}$, which means that $$f(y) = c\lfloor y + \tfrac{1}{2} \rfloor,$$for an arbitrary real constant $c$. It is easy to check that these work.