Croatian Mathematical Competitions

If there are two equal numbers between the given numbers, we can choose those two. So we assume that all the numbers are distinct and without loss of generality we assume that $x_1 > x_2 > \dots > x_n$. Since the total sum of the numbers is zero there exists a positive integer $k$ ($1 < k < n$) such that $x_k > 0 > x_{k+1}$. We have:
$$|x_1| + |x_2| + \dots + |x_k| = |x_{k+1}| + |x_{k+2}| + \dots + |x_n|.$$
Now we assume that there do not exist $i$ and $j$ with the required property. For $i = 1, 2, \dots, k-1$ we have:
$$\frac{|x_i|}{|x_{i+1}|} > 1 > \frac{1}{2},$$
and hence:
$$\frac{x_i}{x_{i+1}} > 2.$$
It follows that:
$$x_i < \frac{1}{2}x_{i-1} < \frac{1}{2^2}x_{i-2} < \dots < \frac{1}{2^i}x_1, \text{ for } i = 1, 2, \dots, k-1.$$
Similarly, we prove that for negative numbers the following inequalities hold:
$$|x_j| < \frac{1}{2^{n-j}}|x_n|, \text{ for } j = k, k+1, \dots, n.$$
Finally, we have:
$$\begin{aligned} |x_1| &< |x_1| + |x_2| + \dots + |x_k| = |x_{k+1}| + |x_{k+2}| + \dots + |x_n| \\ &< |x_n| \left( 1 + \frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2^{n-k+1}} \right) < 2|x_n|. \end{aligned}$$
Analogously, we get:
$$\begin{aligned} |x_n| &< |x_k| + |x_{k+1}| + \dots + |x_n| = |x_1| + |x_2| + \dots + |x_k| \\ &< |x_1| \left( 1 + \frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2^k} \right) < 2|x_1|. \end{aligned}$$
Therefore $\frac{1}{2} \le \left|\frac{x_1}{x_n}\right| \le 2$.