The sum does not vanish when $N$ is odd. I will elaborate on François Brunault's idea.
Assume that $N$ is odd, and let $S$ be the sum in question. Working in the ring $\mathbb{Z}[z]$, we see that $S^{16}\equiv T\pmod{2}$, where $$T:=\sum_{k=0}^{N-1}e^{16\pi i(2k^2+k)/N}=e^{-2\pi i/N}\sum_{k=0}^{N-1}e^{2\pi i(4k+1)^2/N}=e^{-2\pi i/N}\sum_{\ell=0}^{N-1}e^{2\pi i\ell^2/N}.$$ The $\ell$-sum is a classical Gauss sum whose fourth power is known to be $N^2$. Therefore, $S^{64N}\equiv T^{4N}=N^{2N}\equiv 1\pmod{2}$, and hence $S\neq 0$ as claimed.
