9.1. What is the largest number of different natural numbers that can be chosen so that the sum of any three of them is a prime number?
Solution. One example of four numbers that satisfy the condition of the problem is $1,3,7,9$. Indeed, the numbers $1+3+7=11, 1+3+9=13, 1+7+9=17, 3+7+9=19$ are prime.

Suppose it was possible to choose five numbers. Consider the remainders of these numbers when divided by 3. If there are three identical remainders among them, then the sum of the corresponding numbers is divisible by 3. If there are no three identical remainders, then each of the remainders 0, 1, or 2 must be present. Then the sum of three numbers with different remainders when divided by 3 is divisible by 3. This sum is not equal to 3, since all numbers are different and natural. Therefore, this sum is a composite number.

There are other examples as well, such as $3,7,9,31$, etc.

Answer: 4 numbers

Criteria. 7 points - correct and justified solution; 5 points - correct estimate without example; 2 points - correct answer and example provided; 0 points - only the correct answer provided.