Does a connected manifold with vanishing Euler characteristic admit a nowhere-vanishing vector field? A version of the "hairy ball" theorem, due probably to Chern, says that the Euler-characteristic of a closed (i.e. compact without boundary) manifold $M$ can be computed as follows.  Choose any vector field $\vec v \in \Gamma(\mathrm T M)$.  If $p$ is a zero of $\vec v$, then the matrix of first derivatives at $p$ makes sense as a linear map $\partial\vec v|_p : \mathrm T_p M \to \mathrm T_p M$.  By perturbing $\vec v$ slightly, assume that at every zero,  $\partial\vec v|_p$ is invertible.  Then $\chi(M) = \sum_{\vec v(p)=0} \operatorname{sign}\bigl( \det \bigl(\partial\vec v|_p\bigr)\bigr)$.
I am curious about the following potential converse: "If $M$ is closed and connected and $\chi(M) = 0$, then $M$ admits a nowhere-vanishing vector field."
Surely the above claim is false, or else I would have learned it by now, but I am not sufficiently creative to find a counterexample.  Moreover, I can easily see an outline of a proof in the affirmative, which I will post as an "answer" below, in the hopes that an error can be pointed out.  Thus my question:

What is an example of a compact, connected, boundary-free manifold with vanishing Euler characterstic that does not admit a nowhere-vanishing vector field?  (Or does no such example exist?)

 A: Yes, if M is closed and connected and χ(M)=0, then M admits a nowhere-vanishing vector field.
Start with generic vector field, it has zero of index $\pm 1$.
Two zeros of opposite sign can kill each other (maybe it is called Whitney trick?).
So you get a field with zeros of the same sign.
The result follows since the sum of the indexes is the Euler characteristic.
A: That a compact manifold M with vanishing Euler characteristic has a nonvanishing vector field was proved by Heinz Hopf,  Vektorfelder in Mannifaltigkeiten, Math. Annalen  95 (1925), 340-367.  A pretty convincing "intuitive proof" was outlined by Norman Steenrod in his book Fibre Bundles (Theorem 39.7), using a smooth triangulation and obstruction theory.  For a complete proof he refers to page 549 of the 1935 book Topologie by P. Alexandroff and H. Hopf
A: Here is an outline of a proof that a compact manifold $M$ with vanishing Euler characteristic has a nonvanishing vector field.  I'll post it as an "answer" to provide a convenient place for comments where errors might be pointed out (or, you know, a reference given that "that's the proof that ABC used in their paper LMNOP").
Call a vector field $\vec v$ with only regular zeros (i.e. at every $p$ with $\vec v|_p = 0$, the linear map $\partial \vec v|_p : \mathrm T_p M \to \mathrm T_p M$ is invertible) divergence-free if for every zero $p$, $\partial \vec v|_p$ has only real eigenvalues.  A zero $p$ of a divergence-free vector field $\vec v$ has a Morse index $\mu(p) = \#\lbrace$negative eigenvalues of $\partial\vec v|_p\rbrace$.  Of course, $\operatorname{sign}(\det(\partial \vec v|_p)) = (-1)^{\mu(p)}$.  By choosing a Morse function and a metric, our manifold $M$ certainly has a divergence-free vector field.
Pick some vector field $\vec v$ on $M$, and some little neighborhood without any zeros.  I claim I can modify $\vec v$ by some vector field with compact support in that neighborhood to introduce two new zeros, with Morse indexes $\mu$ and $\mu+1$, for any $\mu = 0,\dots,\dim M - 1$.  In one dimension, which is trivial: in a neighborhood, $\vec v = \nabla(x^3+x)$ for some coordinate $x$, and I can perturb this to $x^3 - x$.  In higher dimensions, it is not much harder, and I could probably write out formulas is necessary.
Conversely, and here's the crux of the argument, where I'm not sure it's correct: Suppose I have a divergence-free vector field $\vec v$, with nearby zeros at consecutive Morse index.  Then I think I can cancel them, by undoing the insertion step in the previous paragraph.
If so, then choose any divergence-free vector field $\vec v$ on $M$.  Since $M$ by assumption has vanishing Euler characteristic, $\vec v$ has the same number of zeros with odd Morse index as with even Morse index.  If $\vec v$ has no zeros, we're done; otherwise, choose two with Euler characteristics $\mu$ and $\mu+k$, for $k\geq 1$ odd.  By assumption, $M$ is connected; choose a simply path between the two zeros, and a small neighborhood thereof.  Now, along this path, insert in pairs zeros with Morse index $\mu +1$, $\mu + 2$, ... $\mu + (k-1)$.  Now cancel the zeros in pairs but this time cancel the zero with Morse index $\mu$ with the new one with index $\mu + 1$, and so on.  After all this, you end up with a vector field with two fewer zeros than you started with.
Then the result follows by induction, provided the crux in the 4th paragraph is correct.
