Ok, so we if we delete all the even or odd numbers (25 numbers), all the sums will be even so they will all be composite. Also, for a number like 47 we see 47 = 46 + 1, 47 = 45 + 2 etc. we have to make sure we don't have these pairs on the brownboard so we have to delete at least 23 numbers. If we try to prove we need less than 25 numbers we can go this way: Let's delete any 23 numbers previously shown. Our suffers will come here: We didn't delete 47, 48, 49 and 50, and 97 = 50 + 47 97 = 48 + 49 So we have to delete extra 2 numbers, so 25 deletations (I know this isn't a word but I like to use it) is the minimum we can find. Should be correct and should be quite easy to understand for beginners like me.

To rephrase @above's solution, we need to delete at least 23 numbers from the set {1, 2, 3, ..., 46} (otherwise some pair of them sums to 47), and we also need to delete at least 2 numbers from {47, 48, 49, 50} (otherwise some pair of them sums to 97). Then a minimum 25 deletions are required. We can attain 25 by deleting all of the odd numbers, or all of the even numbers. I also agree with $\boxed{25}$ deletions.

scrabbler94 wrote: To rephrase @above's solution, we need to delete at least 23 numbers from the set {1, 2, 3, ..., 46} (otherwise some pair of them sums to 47), and we also need to delete at least 2 numbers from {47, 48, 49, 50} (otherwise some pair of them sums to 97). Then a minimum 25 deletions are required. We can attain 25 by deleting all of the odd numbers, or all of the even numbers. I also agree with $\boxed{25}$ deletions. Specifically we need to delete $(49,50)$ or $(50,48)$ or $(49,47)$ in the second case

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: $(1, 2),(3, 4),(5, 6),(7, 10),(8, 9),(11, 12),(13, 16),(14, 15),(17, 20), (18, 19),(21, 22),(23, 24),(25, 28),(26, 27),(29, 30),(31, 36),(32, 35), (33, 34),(37, 42),(38, 41),(39, 40),(43, 46),(44, 45),(47, 50),(48, 49)$. Since at least one number from each pair has to be deleted, the minimal number is $25$.

We can also take the sets (50, 3), (49, 4), (48, 5), ....., (27, 26) and (1, 2) So the sum of the first 24 sets is 3 and the last set in 3. Thus, at least one number fron each set should be deleted and we get the minimal number: 25