Note: This problem does not require the "distinct" condition. This solution proceeds for the case of "5 positive primes".

Notice that in the numerical example, of the primes picked, two are congruent to $1$ mod $3$ and two are congruent to $2$ mod $3$ . This is not a coincidence. In fact, every example of 4 primes where the sum of any 3 are prime must satisfy this condition.

As such, this suggests that we look at the values of the primes modulo 3. If we choose five primes, there are the following cases:

Case 1: $3$ is not one of our primes. Let our primes be $p_1,p_2,p_3,p_4,p_5$ . All our primes must either be congruent to $1$ mod $3$ or $2$ mod $3$ . By the pigeonhole principle, three of our five primes must therefore be congruent to each other mod $3$ -- WLOG say $p_1\equiv p_2\equiv p_3\mod 3.$ Then $p_1 + p_2 + p_3 \equiv 0\mod 3$ and, since this sum is greater than three, it must be composite.

Edit: As noted by @Jacob Swenberg, Case 2 can be simplified using the given condition that all the primes are distinct.

Case 2: $3$ is one of our primes. If three of our primes are $3$ , then their sum is $9$ . So assume that at most two of our primes are three. Let our primes be $3,p_2,p_3,p_4,p_5$ , with the possibility that $p_2=3$ . Note that $p_3,p_4,p_5$ all must must be congruent to $1$ mod $3$ or $2$ mod $3$ . If they are all congruent to each other, the argument from Case 1 shows that their is not prime. If they are not all congruent, WLOG say $p_3\equiv 1\mod 3$ and $p_4\equiv 2 \mod 3$ . Then $3 + p_3 + p_4 \equiv 0\mod 3$ and, since this sum cannot be 3, it must be composite.

Lemma: There are no integer solutions to $a+b+c = 5, 0 \leq a, b, c, < 3, abc = 0$ .

Proof: Suppose such a solution exists. WLOG, $a = 0$ . Then, we have $5 = b + c \leq 2 + 2 = 4$ , which is a contradiction.

Lemma: Given any 5 integers, there exist 3 whose sum is a multiple of 3.

Proof: Suppose not. There are 5 integers such that the sum of any 3 integers is not a multiple of 3.
Let $a, b, c,$ represent the number of integers that leave a remainder of 0, 1, and 2 respectively when divided by 3.
Since no 3 sum to a multiple of 3, this means that $0 \leq a, b, c < 3$ otherwise we can take the sum of those 3.
If $abc \neq 0$ , then each of them is at least 1, and the sum of 1 number from each group will give us a multiple of 3. Thus $abc = 0$ .
However, this contradicts the previous claim.

Corollary: Given any 5 integers, each of which is at least 2, there exist 3 whose sum is composite.

Proof: By the 2nd lemma, there exist 3 whose sum is a multiple of 3.
Since each of the integers is at least 2, their sum is at least 6. Thus, their sum is not 3.
Hence, their sum is a composite number.

No. '5 primes such that the sum of ANY three is also a prime.' The real problem is not picking the primes. The problem is 'sum of ANY three' still being prime, and that sum not being divisible by 3, specifically.

Let's build a set of 5 primes noting only their modulo, base 3. We'll pick: +1, +1, -1, -1. So far, okay. But now we're stuck. If we pick another +1, the three +1's added together form a number bigger than 3 with a modulo 3 = 0 so NOT prime. If we pick another -1, the three -1's have the same problem. So we try 3, itself, with modulo 0. Then the choice of a +1 and a -1 also lead to a number with modulo 0, but greater than 3, so not prime. So it doesn't matter how you goose the primes, the problem is you can't pick FIVE of them such that ANY three sum to a prime.

Well, there is one very simple restriction I can add. One of the primes in any set of 5 would be 2.

One of the 5 primes is 2. There must be cases involving (a+b) + 2, which always equals (even Number) plus 2 and is thus never divisible by 2.

If we add four prime numbers, our result will always be an even integer and thus not a prime number. However, there is one exception, which occurs when one of the chosen primes is 2. Hence, in order to make our result odd, 2 must be one of the primes. But, as we have to leave it out once in order to try every possible combination of the five numbers, the tim we leave it out, the result must necessarily be an even number and can't be prime.