The natural numbers from 1 to 50 are written down on the blackboard. At least how many of them should be deleted, in order that the sum of any two of the remaining numbers is not a prime?
Notice that if the odd, respectively even, numbers are all deleted, then the sum of any two remaining numbers is even and exceeds 2, so it is certainly not a prime. We prove that 25 is the minimal number of deleted numbers. To this end, we group the positive integers from 1 to 50 in 25 pairs, such that the sum of the numbers within each pair is a prime:

$$
\begin{aligned}
& (1,2),(3,4),(5,6),(7,10),(8,9),(11,12),(13,16),(14,15),(17,20) \\
& (18,19),(21,22),(23,24),(25,28),(26,27),(29,30),(31,36),(32,35) \\
& (33,34),(37,42),(38,41),(39,40),(43,46),(44,45),(47,50),(48,49)
\end{aligned}
$$

Since at least one number from each pair has to be deleted, the minimal number is 25.