Let $M$ be a smooth manifold (not necessarily orientable) and let $N=\partial M$. Is $N$ necessarily orientable?
I have no particular reason to believe that this is the case, but I wasn't able to come up with a counterexample either.
I'm a bit rusty on this but would not something like $M\times [0,1)$ be a counterexample? (where $M$ is the Möbius band)
The example given by azarel is a non-compact one. Here is a compact example.
We know how to make Klein bottle (i.e. we fibre a circle over a circle in a way). If we replace the fibre by disc then we get a manifold whose boundary is Klein bottle, i.e. the boundary is non-orientable.
The general theorem is that a manifold is a boundary if and only if all of its Stiefel-Whitney numbers are zero. This can happen for both orientable and non-orientable manifolds. The Klein bottle is the simplest example of a non-orientable manifold that is a boundary.
Here is the proof that all of the Stiefel-Whitney numbers are zero for the Klein bottle.
Since the Klein bottle is a 2 dimensional surface it has only two Stiefel-Whitney numbers, the square of its first Stiefel-Whitney class and its second Stiefel Whitney class both evaluated on the mod 2 fundamental cycle.
If one thinks of a Klein bottle as a circle bundle over a circle, then the first Stiefel Whitney class is the intersection number with a fiber circle. Its self intersection number is zero. This can all be easily seen by drawing a Klein bottle as the fundamental domain of the action of the group of isometries of the plane generated by the standard lattice together with the map (x,y) -> ( x + 1/2, -y) The fiber circles are the projections of straight lines that are perpendicular to the x axis.
So the square of the first Stiefel-Whitney class is zero.
The second Stiefel Whitney class is zero because the Klein bottle has an everywhere non-zero vector field. One can also show that it is the square of the first Stiefel-Whitney class, and therefore zero,because the classifying map of the tangent bundle can be factored through the infinite real projective space.
Therefore the Klein bottle is a boundary by Thom's theorem.
But the reply above, that one can extend the fiber circle to a disk is true and this shows that the Klein bottle is a boundary directly. If you construct a Klein bottle from a cylinder by attaching the bounding circles with a reflection, then this map clearly extends to the solid cylinder.
BTW: The first Z/2 cohomology of the Klein bottle is not generated by the first Stiefel Whitney class. Its first cohomology is isomorphic to Z/2 x Z/2.
Interestingly, the square of the other cohomology class, the one which is not the first Stiefel-Whitney class is not zero. This class is the intersection number with the equatorial circle of the Klein bottle. Its self-intersection number, and therefore the square of the cohomology class, is not zero. This follows from the Intermediate Value theorem.