So people I ask say this:

    Pick a door, the host discards 98 doors
    where the car isn't.  There are now two
    doors left, so it's 50:50.

Actually, I'm with them (except for the 50:50 bit). I don't find the 100 door version any more convincing than the 3 door version. Under the usual assumptions the reveal of the other doors gives no information about the one you picked, so that will always remain 1/N. The remaining door will therefore be (N-1)/N, which is bigger if N>2. So switch.