There exists $2n-1$ pairs $(a_i, a_j)$ s.t. $a_i+a_j\geq 0$. Let $a_1, a_2,..., a_{2n}\in \mathbb {R} $ s.t. $a_1+a_2+...+a_{2n}=0$. Show that there exists $2n-1$ pairs  $(a_i, a_j)$ s.t. $a_i+a_j\geq 0$. 
My idea: I consider that $a_1,..., a_k\geq 0$ and $a_{k+1},..., a_{2n}<0$(I put them the sign minus).
Then there are $\frac {k (k-1)}{2} $ pairs. Also $a_1+...a_k=a_{k+1}+...+a_{2n} $. But I am stuck.
 A: Motivation: There are
$$
 \binom{2n}{2} = (2n-1)n
$$
pairs $(a_i, a_j)$ with $1 \le i < j \le 2n$. One can arrange the corresponding sums $a_i + a_j$ in $2n-1$ rows with $n$ entries in per row, so that in each row the entries add up to $a_1 + \ldots + a_{2n} = 0$.
Then every row must have at least one non-negative entry, that makes $2n-1$ sums $a_i + a_j \ge 0$, and the statement is proved.
Examples:
$n=2$:
$$
\begin{array}{ccc}
a_1 + a_2 &           & a_3+a_4 \\
a_1 + a_3 & a_2 + a_4 \\
a_1 + a_4 & a_2 + a_3
\end{array}
$$
$n=3$:
$$
\begin{array}{ccccc}
a_1 + a_2 &           & a_3 + a_4 &           & a_5 + a_6 \\
a_1 + a_3 & a_2 + a_5 &           & a_4 + a_6 \\
a_1 + a_4 & a_2 + a_6 & a_3 + a_5 \\
a_1 + a_5 & a_2 + a_4 & a_3 + a_6 \\
a_1 + a_6 & a_2 + a_3 &           & a_4 + a_5
\end{array}
$$
Proof of the general case: (With the great help of Batominovski!)
Consider the complete graph $K_{2n}$ with vertices $1, \ldots, 2n$. According to Wikipedia: Graph factorization, it has a “1-factorization”, i.e. the edges can be partitioned into disjoint “1-factors” or “perfect matchings.”
Each perfect matching consists of $n$ edges without common vertices
$$
 (i_1, j_1), (i_2, j_2), \ldots , (i_n, j_n)
$$
and since the sum of all $a_k$ is zero, there must be (at least) one vertex $(i_k, j_k)$ such that $a_{i_k} + a_{j_k} \ge 0$.
The total number of edges in $K_{2n}$ is $\binom{2n}{2} = (2n-1)n$, so that the factorization consists of $2n-1$ perfect matches, and we have indeed $2n-1$ distinct vertices $(i_k, j_k)$ with  $a_{i_k} + a_{j_k} \ge 0$.
