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?
1. **Identify the problem**: We need to delete numbers from the set $\{1, 2, 3, \ldots, 50\}$ such that the sum of any two remaining numbers is not a prime number.

2. **Prime sum condition**: For any two numbers $a$ and $b$ in the set, $a + b$ should not be a prime number.

3. **Consider the sum of pairs**: 
   - The smallest prime number is 2, but since we are dealing with natural numbers from 1 to 50, the smallest possible sum is $1 + 1 = 2$.
   - The largest prime number less than or equal to 100 (since $50 + 50 = 100$) is 97.

4. **Strategy for deletion**:
   - We need to ensure that no two remaining numbers sum to any prime number between 2 and 97.
   - Consider the set $\{1, 2, 3, \ldots, 46\}$. If we do not delete at least 23 numbers from this set, there will be at least one pair $(a, b)$ such that $a + b = 47$ (since 47 is a prime number).
   - Similarly, consider the set $\{47, 48, 49, 50\}$. If we do not delete at least 2 numbers from this set, there will be at least one pair $(a, b)$ such that $a + b = 97$ (since 97 is a prime number).

5. **Minimum deletions**:
   - From the set $\{1, 2, 3, \ldots, 46\}$, we need to delete at least 23 numbers.
   - From the set $\{47, 48, 49, 50\}$, we need to delete at least 2 numbers.
   - Therefore, the total minimum deletions required are $23 + 2 = 25$.

6. **Verification**:
   - Deleting all odd numbers or all even numbers from the set $\{1, 2, 3, \ldots, 50\}$ will ensure that the sum of any two remaining numbers is not a prime number.
   - For example, deleting all odd numbers leaves us with the set $\{2, 4, 6, \ldots, 50\}$, and the sum of any two even numbers is even, which cannot be a prime number (except for 2, but 2 is not possible as a sum of two distinct even numbers).

Conclusion:
The minimum number of deletions required is $\boxed{25}$.