Each of the 52 integers can be expressed as 100a + b, where b can range from -49 to 50. That means that the absolute value of b can take only 51 possible values (0 to 50), so at least two of our 52 integers must share one such value. If these two integers have opposite signs associated with b, then the sum of those integers is divisible by 100; if they have the same sign, then their difference is divisible by 100.

I found this in “The Olympiad Corner” in the September 2003 issue of Crux Mathematicorum; reportedly it appeared originally in the Russian Mathematical Olympiad, but the date isn’t known.