For a smooth map $f:M\to N$ from an orientable closed surface $M$ to a non-orientable closed surface $N$, we define its parity (also called modulo 2 degree, and denoted $\deg_2(f)$) as the parity of the number of preimages of any regular value of $f$.
By a geometric construction, I am able to convince myself that in fact $f$ is even, but I suspect that there exists an argument of algebraic topology to show this more easily.
Avoiding the construction, I can say the following:
If there exists an odd map $f:M\to N$, then there exists an odd map $gf:M\to\mathbb P^2(\mathbb R)$, since for any non-orientable $N$ we can construct an odd map $g:N\to\mathbb P^2(\mathbb R)$ by collapsing to a point the complement of a tubular neighbourhood of an orientation reversing loop. So we can assume that $N=\mathbb P^2$, which has odd Euler characteristic. This is interesting because $M$ has even characteristic.
If $f$ is a local homeomorphism, then it's a covering map, so it factors via the orientation covering of $N$, which is even. So in this case it's easy to show that $f$ is even.
Of course, it would also be good to know what happens in greater dimensions. I think that a similar construction shows again that the map must be even.
And although it really isn't necessary for the construction, I would like to know what happens when you collapse to a point the border of a compact manifold-with-border. When can I say that I get a topological manifold? In this case I collapsed a tubular neighbourhood of an orientation-reversing curve. In even dimensions I obtain the projective space, but in odd dimensions I have no idea. I can't obtain the projective space because I can't obtain an orientable manifold.