Let the numbers be $a_1\le a_2\le \cdots \le a_i$. Notice on the one hand $a_{12}+\cdots+a_{21} \le a_1+\cdots+a_{11}$. On the other hand $a_{12}+\cdots+a_{21} \ge a_2+\cdots+a_{11}$. Consequently, $a_1\ge 0$, which concludes the proof.