The sum does not vanish. I will generalize and elaborate on François Brunault's idea.

Let us write $N=2^mM$, where $m\in\mathbb{Z}_{\geq 0}$ and $M$ is odd. Let $S$ be the sum in question. Working in the ring $\mathbb{Z}[z]$, and squaring $S$ multiple times, we see by a familiar level raising argument that $S^{2^{m+1}}\equiv 2^m T\pmod{2^{m+1}}$, where
$$T:=\sum_{k=0}^{M-1}e^{2\pi i(2k^2+k)/M}.$$
It suffices to show that $T^{2M}=(-1)^{(M-1)/2}M^M$, because then $(S^{2^{m+1}}/2^m)^{2M}\equiv 1\pmod{2}$ is a consequence. Let us write
$$w:=1+i\qquad\text{and}\qquad f(z):=e^{2\pi i(2z^2+z)/M}.$$ We shall integrate on translates of the slanted line $w\mathbb{R}$, upwards. An application of the residue theorem shows that
$$S=\int_{-1/2+w\mathbb{R}}\frac{f(z+M)-f(z)}{e^{2\pi i z}-1}\,dz=
\int_{-1/2+w\mathbb{R}}f(z)\,(1+e^{2\pi iz}+e^{4\pi iz}+e^{6\pi iz})\,dz.$$
In the last integral, we can shift the contour to $w\mathbb{R}$, and then after writing $z=wt$ (where $t$ increases from $-\infty$ to $\infty$), it is straightforward to evaluate the four pieces coming from $1$, $e^{2\pi iz}$, $e^{4\pi iz}$, $e^{6\pi iz}$. The final result is
$$T=\exp\left(\frac{\pi i}{4}-\frac{\pi i}{4M}\right)\sqrt{M}.$$
Therefore, $T^{2M}=(-1)^{(M-1)/2}M^M$, and we are done.