This is not an answer, but a derivative question worth pursuing. I put it here in hopes someone can complete it to an answer which solves the posted question.
Let p be a prime dividing n+1. Divide a given set A of integers into residue classes mod p, so there are a_j many members of A equal to j mod p. When I feel up to it, I will write out the equations for how many of the sums of unordered pairs from A add up to a number which is k mod p. The upshot is that for a set A to satisfy the required conditions mod n(n+1)/2, it needs to satisfy an equidistribution system of equations mod p for each such p. In other words, if S_a is the count of such sums from A with sum equal to c mod p, then S_c=S_b for c different from b mod p. We have p choose 2 equations of a form like $$ \sum_{j+k=c \bmod p, j \lt k} a_ja_k + \sum_{j+j=c \bmod p} (a_j^2 + a_j)/2= \sum_{j+k=b \bmod p, j \lt k} a_ja_k + \sum_{j+j=b \bmod p} (a_j^2 + a_j)/2$$.
The question now is are there any tuples of nonnegative integers $a_j$ whose sum is $n$ and which satisfy the above system? For n=2 and p=3, we have (0,1,1). Are there any others?
It is easy to verify (by running through 3-partitions of 5) that for n=5 and p=3, there are no 5- sets A which sums are equidistributed mod 3.
Gerhard "Is This In The Literature?" Paseman, 2018.06.19.