Three of the foregoing six sums consist of only positive integers. Any sum of consecutive positive integers (a+1) + (a+2) + ... + (a+n) is unchanged by placing the sum (-a) + (-a+1) + ... + (a-1) + a in front of it. Therefore, the three "positive only" sums ensure that three more possibilities are feasible.
Consider the factors of 30: 1 x 30, 2 x 15, 3 x 10, and 5 x 6. All odd factors greater than 1 produce consecutive integral sums, as follows. Take 3 x 10, for example. Three consecutive integers with 10 in the middle will add to 30, which produces 9 + 10 + 11.
Observe that 30 = 0.5 x 60. The consecutive integers 0 and 1 average 0.5. Another sum can be formed by taking thirty pairs of integers each totaling 1. The sum is given by (-29) + (-28) + (-27) + ... + 28 + 29 + 30. The result is actually the sum based on putting integers in front of 30, which comes from 30 x 1. Note that 1.5 x 40, 2.5 x 24, and 7.5 x 8 correspond to some of the other foregoing sums.