9.5. For what minimum $\boldsymbol{n}$ in any set of $\boldsymbol{n}$ distinct natural numbers, not exceeding 100, will there be two numbers whose sum is a prime number?
Answer. $\boldsymbol{n}=51$.

Solution. The sum of two even natural numbers is always even and greater than two, hence it cannot be a prime number. Therefore, the example of a set of all fifty even numbers not exceeding 100 shows that the minimum $\boldsymbol{n}$ is not less than 51.

On the other hand, let's divide all natural numbers from 1 to 100 into 50 pairs, the sum of the numbers in each of which is the prime number 101: 1 and 100, 2 and 99, $\ldots, 50$ and 51. If the selected set contains no fewer than 51 numbers, then by the pigeonhole principle, at least two of them will fall into one pair, and their sum will be a prime number.

Grading criteria. Example showing that $\boldsymbol{n}$ is greater than 50: 2 points. Proof that in any set of 51 distinct natural numbers not exceeding 100, there will be two whose sum is a prime number: 5 points.

Idea of the example, but not explicitly provided +1 point.