Well, if it were N*lambda/2, it would be the same as zero (i.e., including N = 0). The segments would simply be in parallel, so the impedance would stack up, but not the poles.
With lambda/4, each stub is a transformer, that is itself transformed. A stub is open at the end, so it acts as a short at the junction. Which acts like an open, 1/4 wave from that; but that next node is also shorted by another stub, and so on.
So the whole thing acts like a thick brick wall at the right frequency.
If the spacing were any other (non integral) length, the transformation would be incomplete, and reactive; all the stubs would have to be weird, and the frequency response would simply be wrong.
A more accurate explanation requires understanding each trace length as a resonator itself; you get coupling from one to the other and so on. The coupling factor, and therefore the bandwidth (given the limitations that the maximum bandwidth has to be fractional, because the stubs and segments still need to look largely 1/4-wave-ish for it to work), is determined by the ratio of impedances. Thus you get fatter traces acting to short out the signal, and narrower traces acting to open up the signal, with the geometric average of those two impedances being equal to the system impedance Zo = sqrt(Zfat * Zthin).
(The coupling can, of course, be by proximity rather than direct connection, and hence you get narrow band filters made from 1/2 wave strips.)
Also... isn't that a lowpass filter? Well, I suppose it'll have harmonic pass bands, so it would be a band stop of sorts. But the other band edges may not be as well controlled as the fundamental cutoff edge. Dunno, depends?
Tim