If the Fourier transform $F(k)$ of $f(x)$ vanishes outside of the interval $(-1,1)$ then, by virtue of <A HREF="https://en.wikipedia.org/wiki/Poisson_summation_formula">Poisson summation,</A>
$$\sum_{n=-\infty}^\infty f(x+n)=\sum_{n=-\infty}^\infty F(n)e^{2\pi inx}=F(0)$$ independent of $x$.

An example is $F(k)=k^2-a^2$ for $|k|<a$ and $F(k)=0$ for $|k|>a$, with $a<1$. Then 
$$f(x)=\frac{ a k \cos a k-\sin a k}{2\pi k^3}$$
and $\sum_n f(x+n)=-a^2$.