If a manifold M has zero Euler characteristic, there is a non-vanishing vector field on it There is hint: if M has isolated singular points, find a diffeomorphism to make these singular points in a any neighborhood which you want. How can we do next?
 A: // Let me begin with an obstruction-theoretic solution — in hope that someone might find it of interest.
(For simplicity, let $\pi_1(M)=0$.) Non-vanishing vector field is a section of the spherization of $TM$, a bundle with fiber $S^{n-1}$. Obstructions to finding a section of a bundle with fiber $S^{n-1}$ lie in groups $H^k(M;\pi_{k-1}S^{n-1})$. These groups are trivial for $k<n$ (since coefficients are trivial) and for $k>n$ (since $M$ is $n$-dimensional). So the only non-trivial obstruction is the principal obstruction $\chi\in H^n(M;\pi_{n-1}(S^{n-1}))=\mathbb Z$.
And it's not hard to show, that it coincides with Euler char (indeed, the value of the obstruction on an $n$-cell is the degree of vector field on the bounding sphere — which coincides with the sum of indices of singular points of an extension of the field inside the cell).
// Reference (obstruction-theoretic approach to char. classes): Milnor-Stasheff, section 12.

This also explains, how to solve the problem directly (actually, it's the same solution in slightly different language). Take any vector field $v$ on $M$, and choose some sphere, containing all singular points. Degree (aka index) of the field on the sphere is exactly $\chi(M)=0$ — which means exactly that there is a non-vanishing extension of the field from the sphere to the ball (coinciding with $v$ on the boundary, but not inside the ball).
