My immediate solution was quite similar to your approach. To look for something different, I thought about how it would work if we were counting down from $M to zero. My current approach puts all the leftovers from repeatedly subtracting $step from $M into the last range:

($N,$M)=@ARGV;
($step,$s)=(int$M/$N,0);
puhs @ranges, [$M-$step,$M],$M-=$step
    while($M and ($M > 2*$step or $M%$step==0));
print 'Last step oversized by ',$M-$step
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This approach seems somewhat shorter, but it certainly is not clearer. Especially the test when to continue is ugly and wrong - if $M is 2*$step -1, you will only get one highly oversized range instead of one undersized and one ideal range. I'm not sure how to fix this.

But this approach reminds me of Bresenham's line algorithm, which more or less has the same problem, to segment a line into roughly equal-sized segments, except that it distributes these segments in a 2-dimensional way. This algorithm likely has no nicer formulation than what you already have though, but it will distribute the oversized ranges much nicer, and the size difference between two ranges will never be larger than 1.