Sure.

Call the sum of all assigned numbers N. No prisoner knows N, but they do know that there are exactly 100 possible last two digits of N, from "00" up to "99". Beforehand, assign a different last digit pair to each prisoner. When the prisoner guesses, he sums everybody else's number, then guesses the number between 1 and 100 that, when added to his partial sum, results in his assigned last two digits.

That's much cleaner than I thought it would be. I was convinced that there was a solution that guaranteed success, since each arrangement of numbers could get mapped to a unique guessing strategy, so if you could make it so that exactly one correct guess per arrangement you're golden. I tried working with a N-wide N-dimensional cube, got it working for N=2, and failed at N=3.

This makes sense, though twhb didn't mention any restriction on the numbers you might guess. Thus the first prisoner could just guess on the entire sum they saw, and only two guesses would be needed.