A probability problem by Paola Ramírez

Make a partition of the set $\left\{ 1,2,...,100 \right\}$ as follows:

$\left\{ 1,100 \right\}$ , $\left\{ 2,99 \right\}$ ,..., $\left\{ 50,51 \right\}$

The sum of any two elements of the same subset is 101, thus, we can choose at most 1 element of each set such that the sum of any two elements, that we have selected, is different to 101. Indeed, we can pick 50 elements, and, by Pigeonhole Principle, the next element we choose will belong to the same set of another element already selected. Therefore we have to pick up at least 51 elements of the initial set.