I understand that the top Stiefel Whitney class is an obstruction for the tangent bundle of a manifold to have a trivial line sub-bundle. I am looking for a counterexample when removing the word "trivial", i.e: A compact manifold $M$ of dimention n such that $w\_n(TM)\neq0$ (or equivalently $\chi(M)$ is odd) and there exists a sub-bundle $\xi\subset TM$ with $\operatorname{rank}(\xi)=1$.


After some thought I understand why one cannot find an example using surfaces, odd-dimensional manifolds or any of $S^n$,$T^n$,$\mathbb{R}P^n$,$\mathbb{C}P^n$. I also understand why manifolds with $H^1(M;\mathbb{Z}\_2)=0$ will not work, and why such a manifold (if it exists) cannot be null-cobordant.


My intuition on 4-manifolds (or god forbid anything higher dimensional) is not good enough to know where to look. Any insight as to where one should look for such a creature (or why the hell it should not exist) will be greatly appreciated.



A Compact Manifold with odd Euler characteristic whose tangent bundle admits a field of lines
I believe there is no example satisfying all your constraints. If I recall (my memory is a little foggy on this) the result likely goes back to Hopf, and one of his variations on the Poincare-Hopf index theorem. This question might be addressed in the Milnor and Stasheff text. Here is one way to argue the point. 


Say the tangent bundle of the manifold $N$ admits a field of lines. Then (at worst) some 2:1 cover of $N$ admits an everywhere non-vanishing vector field. Since the Euler characteristic is the obstruction to such a vector field existing (Poincare-Hopf index theorem) the Euler characteristic of this covering space is zero. But $\chi N$ is a multiple of the euler characteristic of the cover. 


i.e. your assumption that the Euler characteristic is odd excludes the possibility of a 1-dimensional sub-bundle. 


