Call the 5 numbers A, B, C, D and E. We assume these to be integers, since there's no indication otherwise. The sums are A+B, A+C, A+D, A+E, B+C, B+D, B+E, C+D, C+E, D+E. Let's look at the sum zero. To get the sum of two number to be zero, they're either both zero or one is the negative of the other. Let's assume there are no negative numbers, so two numbers must be zero. So A=B=0. 6 of the remaining numbers involve A or B, so the sums would be C, D, E, C, D, E, i.e., three pairs of identical sums. But we only have one pair, so the assumption of no negative numbers is wrong. Therefore, there is at least one negative number in the set and A=-B. The sums involving A and B we can write C+A, C-A, D+A, D-A, E+A, E-A. Do we have 6 sums that could fit these? Yes, we do. The 6 sums would enable us to find A, since, for example, (C+A)-(C-A)=2A=(D+A)-(D-A)=(E+A)-(E-A). The difference between the pairs of sums is 2 for each pair, implying that A=1 and therefore B=-1. The first two pairs are 2 and 4 and 4 and 6, because we have two 4's. If we add the sums in each pair we get 6 and 10. These are respectively 2C and 2D, so C=3 and D=5, making C+D=8, another of the sums. We're left with sums 9, 11, 13, 15, and we haven't found E yet. We need sums of E with the other discovered numbers, E+1, E-1, E+3, E+5 and we have four remaining sums to fit these if possible. The lowest number is 9 which must correspond to E-1. So E=10 and the other sums become 11, 13 and 15, which matches the three remaining sums. So the answer is: the numbers are -1, 1, 3, 5, 10.