Let's take the group of numbers mod 3. There are three congruency classes and if there are three numbers in the same congruency class then their sum is a multiple of 3, and that sum cannot be the prime 3 itself since (1,1,1) is banned by the footnote.
Then we cannot have three of the numbers in the same congruency class mod 3. So there are at most six numbers, two in each class.
Now if there is one number in each of the three classes then that sum is also a multiple of 3.
This then reduces the group of numbers to being in only two of the classes, so now we are down to a maximum of four numbers.
The groups {1,1,5,5} and {1,3,7,9} both have all their three terms sums being prime so the answer is four.
Solution
