Saudi Arabia Mathematical Competitions

Since ${\sum }_{k=1}^n{a}_k=0$, it follows that $$\sum _{k=1}^{n}k{a}_k=\sum _{k=1}^n(k-x)a_k$$ for every $x\in \mathbb{R}$. We obtain $$\begin{array}{rl}\left|\sum _{k=1}^{n}k{a}_k\right|& =\left|\sum _{k=1}^n(k-x)a_k\right|\le \sum _{k=1}^n|k-x|\left|a_k\right|\\ & \le M(x)\sum _{k=1}^n\left|a_k\right|=M(x)\end{array}$$ where $M(x)=\max \{|k-x|:k=1,2,\dots ,n\}$. For $x=\frac{n+1}{2}$ one has $M(x)=\frac{n-1}{2}$, and the conclusion follows.

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Alternative solution.

For each $k=1,\dots ,n$, we have $$\left|a_1+\dots +a_k\right|+\left|a_{k+1}+\dots +a_n\right|\le |a_1|+\dots +|a_n|=1.$$ The relation $\left|a_1+\dots +a_k\right|=\left|a_{k+1}+\dots +a_n\right|$ implies $$\left|a_1+\dots +a_k\right|\le \frac{1}{2},k=1,\dots ,n$$ Now we can write $$\begin{array}{rl}\left|a_1+2a_2+\dots +n{a}_n\right|& \le |a_1+\dots +a_n|+|a_2+\dots +a_n|+\dots +|a_n|\\ & \le \frac{n-1}{2},\end{array}$$ and we are done.