If the Fourier transform $F(k)$ of $f(x)$ vanishes outside of the interval $(-1,1)$ then, by virtue of Poisson summation, $$\sum_{n=-\infty}^\infty f(x+n)=\sum_{n=-\infty}^\infty F(n)e^{2\pi inx}=F(0)$$ 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 $$f(x)=\frac{ a x \cos a x-\sin a x}{2\pi x^3}\;\;\text{and}\;\;\sum_n f(x+n)=-a^2.$$