Poisson summtion formula has several forms. The most well known is that $$\sum_{n\in\mathbb{Z}} f(n)=\sum_{m\in\mathbb{Z}}\hat f(m)$$ with $\hat f(m)$ being the Fourier transform of $f$ at the point $m$. In another form, we know that for every $P>0,$ $$\sum_{n\in\mathbb{Z}} f(nP+k)=\frac 1P\sum_{m\in\mathbb{Z}}e^{\frac{2\pi i km}{P}}\hat f(m).$$ However the last form implies that for every function $$\sum_{n\in\mathbb{Z}}f(n)=\sum_{k=0}^{P-1}\sum_{n\in\mathbb{Z}}f(nP+k)=\sum_p\frac 1P\sum_{m\in\mathbb{Z}}e^{\frac{2\pi i km}{P}}\hat f(m)=\frac 1P\sum_{m\in\mathbb{Z}}\left(\sum _{k=0}^{P-1}e^{\frac{2\pi i km}{P}}\right)\hat f(m).$$ However, the sum in the brackets is $0$ for every $P,m\in\mathbb N$ meaning $\sum f(n)=0$ for every $f$.
What has gone wrong in this calculation? Needless to say $\sum f(n)$ can be of course a non-zero number or eiven a dievregent series.