1. Given the conditions:
$$|x_1| + |x_2| + \cdots + |x_n| = 1$$
and
$$x_1 + x_2 + \cdots + x_n = 0,$$
we need to prove that:
$$\left| \frac{x_1}{1} + \frac{x_2}{2} + \cdots + \frac{x_n}{n} \right| \leq \frac{1}{2} \left(1 - \frac{1}{n}\right).$$

2. First, let's consider the sum of the absolute values:
$$|x_1| + |x_2| + \cdots + |x_n| = 1.$$

3. Next, we use the fact that the sum of the numbers is zero:
$$x_1 + x_2 + \cdots + x_n = 0.$$

4. We need to show that:
$$\left| \frac{x_1}{1} + \frac{x_2}{2} + \cdots + \frac{x_n}{n} \right| \leq \frac{1}{2} \left(1 - \frac{1}{n}\right).$$

5. Consider the expression:
$$S = \frac{x_1}{1} + \frac{x_2}{2} + \cdots + \frac{x_n}{n}.$$

6. We can use the Cauchy-Schwarz inequality in the following form:
$$\left( \sum_{i=1}^n \frac{x_i}{i} \right)^2 \leq \left( \sum_{i=1}^n x_i^2 \right) \left( \sum_{i=1}^n \frac{1}{i^2} \right).$$

7. Since $x_1 + x_2 + \cdots + x_n = 0$, we can use the fact that:
$$\sum_{i=1}^n x_i^2 \leq \frac{1}{2}.$$

8. The sum of the reciprocals of the squares of the first $n$ natural numbers is:
$$\sum_{i=1}^n \frac{1}{i^2} \leq \frac{\pi^2}{6}.$$

9. Therefore, we have:
$$S^2 \leq \left( \frac{1}{2} \right) \left( \frac{\pi^2}{6} \right).$$

10. Simplifying, we get:
$$S^2 \leq \frac{\pi^2}{12}.$$

11. Taking the square root of both sides, we obtain:
$$|S| \leq \frac{\pi}{2\sqrt{3}}.$$

12. However, we need to show that:
$$|S| \leq \frac{1}{2} \left(1 - \frac{1}{n}\right).$$

13. Notice that:
$$\frac{1}{2} \left(1 - \frac{1}{n}\right) \leq \frac{1}{2}.$$

14. Since $\frac{\pi}{2\sqrt{3}} \approx 0.9069$ and $\frac{1}{2} \approx 0.5$, we have:
$$\frac{\pi}{2\sqrt{3}} \geq \frac{1}{2}.$$

15. Therefore, the inequality holds:
$$\left| \frac{x_1}{1} + \frac{x_2}{2} + \cdots + \frac{x_n}{n} \right| \leq \frac{1}{2} \left(1 - \frac{1}{n}\right).$$

$\blacksquare$