jmc

Since $f\left(\frac{1}{2}\right)=\lfloor \frac{1}{2}\rfloor +\frac{1}{2}=\frac{1}{2}$ and $f\left(-\frac{1}{2}\right)=\lfloor -\frac{1}{2}\rfloor +\frac{1}{2}=-\frac{1}{2},$ so if $f$ is either even or odd, it must be odd.

But $f(0)=\lfloor 0\rfloor +\frac{1}{2}.$ Every odd function $f(x)$ satisfies $f(0)=0,$ so $f(x)$ is $\boxed{\text{neither}}.$