SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Denote $$S_x=\left|\sum _{i=1}^x{a}_i-\sum _{i=x+1}^n{a}_i\right|.$$ Obviously $S_0=-S_n$ and $S_{x+1}-S_x=2a_{x+1}$. We may assume that $S_n\ge 0$ and consider $m$ for which $S_m\le 0$ and $S_{m+1}\ge 0$. Then $$\left|S_m\right|+\left|S_{m+1}\right|=\left|S_m-S_{m+1}\right|=2\left|a_{m+1}\right|$$ so either $\left|S_m\right|$ or $\left|S_{m+1}\right|$ is at most $a_{m+1}$ which is at ${\mathrm{most}}_{{\max }_i}\left|a_i\right|=a_k$. This completes the proof.