Solution:

The answer is (C). A number is divisible by $100$ if and only if its decimal representation ends with $00$.

Let us divide the integers $a_{1}, \ldots, a_{n}$ into groups as follows: in a first group we put those whose decimal representation ends with $00$, in a second group those ending with $01$ (or having a single digit and that digit is $1$) or with $99$, in a third those having as last two digits $02$ or $98$ and so on, until we reach the fifty-first group, in which we insert those ending with the digits $50$.

If there are two integers in the same group, then necessarily either their difference or their sum will be divisible by $100$: indeed, if the two integers end with the same group of two digits, then their difference ends with $00$ and is divisible by $100$; if instead they end with different groups of digits, the fact that they are in the same group means that their sum is divisible by $100$. Since we have exactly $51$ groups, $n$ is at most $51$: indeed, if we had $52$ or more integers, at least two would fall into the same group and, for the reasons stated, either their difference or their sum would be divisible by $100$.

On the other hand, it is easy to convince oneself that $100, 101, \ldots, 150$ is precisely a set of $51$ positive integers with the property described in the text, and therefore $51$ is the number required.