Pairs $x_i+x_j$ positive for total positive sum Let $x_1,\dots,x_n\in\mathbb{R}$ be such that $x_1+\dots+x_n>0$. At least how many sums $x_i+x_j$ ($i<j$) must be positive?
It is possible that $x_1=n$ and $x_2=\dots=x_n=-1$, in which case $n-1$ pairwise sums are positive. This should be the best as well. Can we argue something using linear algebra and basis?
 A: Let $f(n)$ be the minimum number of positive sums with $n$ terms.
We have $f(n)=n-2$ for $n=3,5$ and $f(n)=n-1$ otherwise.
Lemma: if $f(n)=n-1$ then $f(n+2)=(n+2)-1$
Proof: We let $x_1<x_2\dots < x_n$. If $x_1+x_{n+2}>0$ we are done.
Otherwise $x_2+x_3+\dots + x_{n+1}>0$ and so there are at least $n-1$ pairs within these number. Clearly this implies $x_{n-1}+x_{n-2}>0$.
Therefore $x_{n+1}+x_n$ and $x_{n+1}+x_{n-1}$ are also positive, and there are at least $n-1$ positive pairs. Notice that for equality to hold we must have $x_{n-3}+x_n\leq 0$ (used later).
Since $f(2)=1$ we have $f(2k)=2k-1$ for all $k$.
We have $f(3)=1$ (take $1,1,-1$).
We have $f(5)=3$ (use the lemma to see $f(5)\geq f(3)+2)$
We have $f(7)=6$, suppose not, let $x_1<x_2\dots <x_n$. By the lemma we must have $x_4+x_7\leq 0$, therefore $x_4\leq 0$ and $|x_4|\geq |x_7|$. from here $|x_1+x_2+x_3+x_4|\geq |4x_4|\geq |3x_7|\geq | x_5+x_6+x_7|\implies x_1+x_2+x_3+x_4+x_5+x_6+x_7\leq 0$.
So $f(7)=6$, we conclude $f(n)=n-1$ for $n\neq 3,5$ and $f(n)=n-2$ for $n=3,5$
A: When $n$ is even: In total there are $\frac{n(n-1)}{2}$ pairs.
these can be split into $(n-1)$ groups of $n/2$ pairs each, such that every number is in exactly one pair of each group.
Here is the way to do it (each color is one group):

It is clear at least one pair per group must have positive sum.
