TITLE: Proving at least 99 pairwise sums of the reals are non-negative
QUESTION [5 upvotes]: Suppose, one hundred real numbers are given and their sum is 0.Then how can I prove that at least 99 of the pairwise sums of these hundred numbers are non-negative?
I tried this:
Let the real numbers be $a_i$,$1\leq i\leq100$.
By the given condition, $a_1+a_2+\dots +a_{100}=0$.Suppose for the sake of contradiction, that at most 98 of the pairwise sums of these given reals are non-negative.I just tried to examine the case when exactly 98 of them are non-negative.
I labelled the sums $S_1$,$S_2,\dots S_{4950}$,WLOG assume that sums $S_1,S_2,\dots S_{4852}$ are negative, the rest are non-negative.
$$S_1+S_2+\dots +S_{4950}=99(\sum_{i=1}^na_i)=0$$
Case I: $S_{4853}+a_{4854}+\dots +a_{4950}=0$.That simply means that $S_1+S_2+\dots +S_{4852}=0$.But, by the given condition, $S_1+S_2+\dots +S_{4852}<0$,a contradiction.After that, I am not able to make any progress.
Can anyone please suggest a more efficient method to solve this problem.I feel this approach will involve enormous case work.

REPLY [7 votes]: Without loss of generality, $a_1\ge a_2\ge \ldots\ge a_{100}$.
From $$0=(a_1+\ldots+a_{50})+(a_{51}+\ldots+a_{100})\le 50a_1+50a_{51},$$ we conclude $a_1+a_{51}\ge0$, hence we have $50$ nonnegative sums $$\tag1 a_1+a_k,\ 2\le k\le 51$$ involving $a_1$.
We may assume that $a_1+a_{100}<0$ as otherwise $(1)$ can be extended to $2\le k\le 100$ and we are done. Therefore $a_2+\ldots+a_{99}=-(a_1+a_{100})>0$ and from
$$0< (a_2+\ldots+a_{50})+(a_{51}+\ldots a_{99})\le 49a_2+49a_{51},$$
we conclude $a_2+a_{51}> 0$, hence we have another $49$ nonnegative (in fact positive) sums
$$\tag2a_2+a_k,\ 3\le k\le 51$$
and together with $(1)$ this produces the required $99$ nonnegative sums.

Warning: Contrary to what it may look like, this method will not guarantee us $50+49+48+\ldots$ nonnegative sums as the first step from $(1)$ to $(2)$ comes into play only if $a_1+a_{100}<0$. In fact, $a_1=99, a_2=\ldots=a_{100}=-1$ shows that $99$ cannot be improved.