A quick comment on the idea of "a very slick solution using the hairy ball theorem". Any such very slick proof will surely only use the fact that the Euler characteristic is non-zero, and so should apply just as well to surfaces of higher genus (at least two). For instance, if I understand it correctly, then Reid's answer above would work just as well on a higher-genus surface, by the Poincare--Hopf Theorem.
But the theorem does not hold on surfaces of higher genus.
What follows would probably be better with pictures, but I'll try to describe it without and hope for the best.
For instance, it's easy to divide a torus into two rectangles such that there are "traffic schedules" with no crashes. (OK, the Euler characteristic of a torus is zero, but bear with me.) (Also, what is to ants as a traffic schedule is to cars?)
Now consider the surface of genus two as an octagon with sides identified, as usual. You can divide this into two rectangles and two pentagons in a way that mimics two copies of the torus picture in the previous paragraph: the surface is the union of two tori with boundary, and each torus is divided into a rectangle and a pentagon. Schedule each torus as before (stretching one of the ants' route over the fifth side of the pentagon). As only one ant from each torus traverses the fifth side of the pentagon, it is easy to arrange that the ants from different tori do so at different times.
Does that make sense, or does anyone want a picture?
EDIT:
Oh, and of course it follows that the theorem is also false on any orientable surface of even higher genus, as they all cover the surface of genus two.
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