Example 2. The sum of the reciprocals of four unequal natural numbers is 1. Find these four numbers.
Let the four numbers be $x, y, z, w$, and $x \geq 1$. Also, when $x \geq 3$, we have
$$
\begin{array}{l}
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{w} \\
\leqslant \frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6} < 1. \text{ Also, when } y \geqslant 6, \text{ we have } \frac{1}{y}+\frac{1}{z}+\frac{1}{w} \leqslant \frac{1}{6}+\frac{1}{7}+\frac{1}{8}<\frac{1}{2}.
\end{array}
$$
Thus, $y=3,4,5$.

(i) When $y=3$, $\frac{1}{z}+\frac{1}{w}=\frac{1}{6}$. By the same discussion, we get $z=7,8,9,10,11$. Solving further, we get $w=42,24,18,15,13 \frac{1}{5}$. Therefore, there are four solutions that satisfy the conditions.

(ii) When $y=4$, $\frac{1}{z}+\frac{1}{w}=\frac{1}{4}$. By the same discussion, we get $z=5,6,7$. Solving further, we get $w=20,12, 9 \frac{1}{3}$. Therefore, there are two solutions that satisfy the conditions.

(iii) When $y=5$, $\frac{1}{z}+\frac{1}{w}=\frac{3}{10}$. By the same discussion, we get $z=4,5,6$. Solving further, we get $w=20, 10, \frac{2}{15}$. However, these three sets of numbers do not satisfy the conditions $x<y<z<w$ and the assumption that they are natural numbers, so they are discarded.

In summary, the original equation has six solutions: $(2,3,7,42)$, $(2,3,8,24)$, $(2,3,9,18)$, $(2,3,10,15)$, $(2,4,5,20)$, $(2,4,6,12)$.