Here is an incomplete proof that $\chi(\tilde{M})=0$ implies that $\chi(M)=0$ for fundamental groups with certain finiteness properties; maybe somebody will come along and finish it off (or find a counter-example)!

The idea is to adapt the usual spectral sequence proof that the Euler characteristic is multiplicative on fibrations. Let $k$ be a field of characteristic zero, and let $\pi=\pi_1(M)$. The Cartan-Leray spectral sequence of the regular cover $\tilde M\to M$ has $E^2 = H_p(\pi;H_q(\tilde M;k))$ and converges to $H_{p+q}(M;k)$. But we can also go one page back and start at $E^1 =F_p\otimes_{k[\pi]} H_q(\tilde M;k)$, where $F_\bullet$ is a free resolution of $k$ by $k[\pi]$-modules. (Reference: K. Brown, Cohomology of groups, VII.5 and VII.7.) Now in a spectral sequence the Euler characteristic of each page is the Euler characteristic of the previous one, so it suffices to check that $\chi(E_1)=0$ when $\chi(\tilde M)=0$.

If the group $\pi$ admits a finite resolution, we can choose each $F_p$ to be a finitely generated free $k[\pi]$ module (I think), so $F_p$ is a direct sum $k[\pi]^{n(p)}$ for some $n(p)$. I think we then get $F_p\otimes_{k[\pi]} H_q(\tilde M;k)\cong k^{n(p)}\otimes_k H_q(\tilde M;k)$ as vector spaces, and the result follows in the usual way.

Here is sketch proof that $\chi(\tilde{M})=0$ implies that $\chi(M)=0$ for fundamental groups with certain finiteness properties.

The idea is to adapt the usual spectral sequence proof that the Euler characteristic is multiplicative on fibrations. Let $k$ be a field of characteristic zero, and let $\pi=\pi_1(M)$. The Cartan-Leray spectral sequence of the regular cover $\tilde M\to M$ has $E^2 = H_p(\pi;H_q(\tilde M;k))$ and converges to $H_{p+q}(M;k)$. But we can also go one page back and start at $E^1 =F_p\otimes_{k[\pi]} H_q(\tilde M;k)$, where $F_\bullet$ is a free resolution of $k$ by $k[\pi]$-modules. (Reference: K. Brown, Cohomology of groups, VII.5 and VII.7.) Now in a spectral sequence the Euler characteristic of each page is the Euler characteristic of the previous one, so it suffices to check that $\chi(E^1)=0$ when $\chi(\tilde M)=0$.

If the trivial module $k$ admits a finite free resolution by $k[\pi]$-modules, we can choose each $F_p$ to be a finitely generated free $k[\pi]$ module, so $F_p$ is a direct sum $k[\pi]^{n(p)}$ for some $n(p)$. I think we then get $F_p\otimes_{k[\pi]} H_q(\tilde M;k)\cong k^{n(p)}\otimes_k H_q(\tilde M;k)$ as vector spaces, and the result follows in the usual way since the Euler characteristic of a tensor product of finite dimensional graded vector spaces is the product of their Euler characteristics.