Kids At An Orchard

The least number of apples they can take is $0+1+2+3+...+14=\dfrac{14\times 15}{2}=105>100$ , so it is not possible.

The first and foremost possibility we have to check is what would happen if one child takes 1 apple, the next one takes two apples, the next one takes 3 apples the next ine takes 4 apples an so on.

We find that $\displaystyle \sum_{n=1}^{13} n = \dfrac{13 × 14}{2} = 91$ .
Then $\displaystyle \sum_{n=1}^{14} n = \dfrac{14 × 15}{2} = 105$ . But there are only 100 apples, so the 14th child cannot take 14 apples. According to question, a child takes a non-negative integer number of apples. So, a child can take 0 apples. So, the only child left is the 15th one. But now only 9 apples are left. So, the 15th child would have to take 9 apples to satisfy the condition of the question. But that would be the same number of apples as taken by the 9th child (i.e. 9 apples). So, there is no possibility of atleast two children taking the same number of apples.

Applying the Pigeonhole principle, there is atleast $[100/15]$ kids who has two or more apples. Done.