Suppose I have a set of such numbers with sum $S$. If I add $1$ to each of the five numbers, I don't change the fact that the condition is achieved, but I change the sum to $S+5$. This means that the only possible answers to the question are the numbers $1$ and $5$. [This observation is not necessary, but is the kind of useful guide to what might be happening and also check on calculations which comes in handy].
I can spot two chains of nine sums which must be different (one given in another answer). The corresponding terms of the chains must then be equal, and that shows that the terms must be in arithmetic progression. Five consecutive terms of an AP have sum divisible by $5$. I suggest trying to find the second chain for yourself.
Then with $a\gt b\gt c \gt d \gt e$ I have both $$2a\gt a+b\gt a+c\gt a+d\gt a+e\gt b+e\gt c+e\gt d+e \gt 2e$$ and $$2a\gt a+b\gt 2b\gt b+c\gt2c\gt c+d\gt 2d \gt d+e \gt 2e$$ whence $$a+c=2b, a+d=b+c, a+e=2c, b+e=c+d, c+e=2d$$ from which $$a-b=b-c, a-b=c-d, [a-c=c-e], b-c=d-e, c-d=d-e$$ showing that there is a common difference - the middle term is not needed for this.