Prove that among any 100 integers there are always two whose difference is divisible by...

Step 1
Let $r_i$ be the remainder when the i-th integer in our list is divided by 99.
There are only 99 conceivable remainders, the numbers 0 to 98.
Step 2
Since there are 100 integers in our list, and only 99 conceivable remainders, there must be two different numbers in our list which have the same remainder. This is a consequence of the Pigeonhole Principle, but the fact is clear even without a name for it.
Finally, if two numbers have the same remainder on division by 99, their difference is divisible by 99.

Step 1
Here's a proof requiring no knowledge of division or remainders. Suppose a counterexample exists, i.e. there are 100 integers ${\mathrm{a}}_1,\dots ,{\mathrm{a}}_{100}$ such that $\mathrm{i}\ne \mathrm{j}\Rightarrow 99\nmid {\mathrm{a}}_{\mathrm{i}}-{\mathrm{a}}_{\mathrm{j}}.$. Then they are distinct, so we may index them so that they're strictly ordered ${\mathrm{a}}_1<{\mathrm{a}}_2<\dots <{\mathrm{a}}_{100}.$. Now if ${\mathrm{a}}_{100}\ge {\mathrm{a}}_1+99$ then by replacing ${\mathrm{a}}_{100}$ by ${\mathrm{a}}_{100}-99$ and reordering we obtain a new counterexample of smaller interval length ($=max-min+1$), since the new sequence has the same min, by ${\mathrm{a}}_{100}-99\ge {\mathrm{a}}_1,$, but has smaller max ${\mathrm{a}}_{99}<{\mathrm{a}}_{100},$, and the new sequence remains a counterexample, since $99\mid {\mathrm{a}}^{\prime }-\mathrm{a}⟺99\mid {\mathrm{a}}^{\prime }-(\mathrm{a}-99).$.
Step 2
Thus the least length counterexample must have length $\le 99,$, hence its 100 integers all lie in the length 99 interval of consecutive integers ${\mathrm{a}}_1,{\mathrm{a}}_1+1,\dots ,{\mathrm{a}}_1+98.$. Therefore, by the pigeonhole (box) principle, two of the integers are equal, a contradiction.