Saudi Arabian 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
$$\begin{aligned} \left|\sum_{k=1}^{n} k a_{k}\right| & = \left|\sum_{k=1}^{n} (k - x) a_{k}\right| \leq \sum_{k=1}^{n} |k - x| \left|a_{k}\right| \\ & \leq M(x) \sum_{k=1}^{n} \left|a_{k}\right| = M(x) \end{aligned}$$
where $M(x) = \max \{|k - x| : k = 1, 2, \ldots, n\}$.
For $x = \frac{n+1}{2}$ one has $M(x) = \frac{n-1}{2}$, and the conclusion follows.

For each $k = 1, \ldots, n$, we have
$$\left|a_{1} + \ldots + a_{k}\right| + \left|a_{k+1} + \ldots + a_{n}\right| \leq |a_{1}| + \ldots + |a_{n}| = 1.$$
The relation $\left|a_{1} + \ldots + a_{k}\right| = \left|a_{k+1} + \ldots + a_{n}\right|$ implies
$$\left|a_{1} + \ldots + a_{k}\right| \leq \frac{1}{2}, \quad k = 1, \ldots, n$$
Now we can write
$$\begin{aligned} \left|a_{1} + 2 a_{2} + \ldots + n a_{n}\right| & \leq |a_{1} + \ldots + a_{n}| + |a_{2} + \ldots + a_{n}| + \ldots + |a_{n}| \\ & \leq \frac{n-1}{2}, \end{aligned}$$
and we are done.