First, we cannot have three numbers in the set of the same parity. If we did, then their three pairwise differences would be even, so would all be $2$, but clearly this is impossible. Thus, our set contains at most two odd and two even numbers, and in particular contains exactly four elements, two odd and two even. Furthermore, the two odd elements are $2$ apart, so exactly one of $2013$ and $2017$ are in the set along with $2015$. For the moment let's assume it is $2013$. Our two even numbers are also $2$ apart, so are either both less than $2013$ or both greater than $2015$. Let us assume the former. Then our set is $a,a+2,2013,2015$ for some appropriate even $a$. Then $2011-a,2013-a,$ and $2015-a$ are all prime, but this can only happen if they are the primes $3,5,7$. Thus, $a=2008$, and we get the set $2008,2010,2013,2015$. In a similar manner we can find exactly four sets, based on choosing either $2013$ or $2017$, and on the evens being higher or lower than $2015$. They are $2008,2010,2013,2015$, $2013,2015,2018,2020$, $2010,2012,2015,2017$, $2015,2017,2020,2022$.

good one