9.3. On the board, there are $N$ prime numbers (not necessarily distinct). It turns out that the sum of any three numbers on the board is also a prime number. For what largest $N$ is this possible
Answer: $N=4$.

Solution. Consider the remainders when the $N$ written numbers are divided by 3. All three remainders cannot occur, because in this case, the sum of three numbers with different remainders will be divisible by 3 (and will be greater than 3), so it will not be a prime number. Therefore, there can be no more than 2 possible remainders. Also, note that there cannot be 3 numbers with the same remainder, because the sum of these three numbers will be divisible by 3 (and will be greater than 3), so it will not be a prime number. Thus, $N$ cannot be greater than $2 \cdot 2=4$.
$N$ can equal 4. For example, if the numbers written on the board are $3,3,5,5$ (sums $11,13)$.

Remark. There are other examples. For instance, $3,3,5,11$ (sums 19, 17, 11). Comment. It is proven that $N$ is not greater than $4-5$ points. An example with 4 numbers is provided - 2 points.