By *symplectic manifold* I mean a pair $(M^{2n},\omega)$ consisting of a smooth, connected, even dimensional manifold and a non-degenerate $2$-form. I am interested in compact, boundarlyess examples where $\chi(M)=0$. If none such exist, can anyone provide a simple proof (understandable to a Topologist who knows a little geometry)?

In case the answer to the question in the title is a quick "yes", I have several follow up questions:

1. What about if we restrict to the case $n=2$ (in which case $M$ would have to be non-simply-connected)?
2. What about if we restrict to closed symplectically aspherical manifolds? [Recall that $(M^{2n},\omega)$ is called *symplectically aspherical* if the symplectic class $[\omega]\in H^2(M;\mathbb{R})\cong \mathrm{Hom}(H\_2(M),\mathbb{R})$ vanishes on the image of the Hurewicz homomorphism $h\colon \pi\_2(M)\to H\_2(M)$.]

Thanks.