12.73 Suppose $x, y, z$ are three distinct natural numbers. Arrange them in ascending order, and the sum of their reciprocals is also a natural number. Find these three natural numbers $x, y, z$.
(Hungarian Mathematical Olympiad, 1918)
[Solution] Let the sum of the reciprocals of these three natural numbers be $a$, and $x<y<z$, then we have
$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=a \text {. }
$$

Since $x, y, z$ are natural numbers and are distinct, we have
$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}<\frac{1}{1}+\frac{1}{2}+\frac{1}{2}=2 \text {. }
$$

Thus,
$$
a<2 \text {. }
$$

Since $a$ is a natural number, we have
$$
a=1 \text {. }
$$

Therefore, the given equation becomes $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$.
$\cdot$ Also, since
$$
\frac{1}{x}<\frac{1}{x}+\frac{1}{y}+\frac{1}{z}<\frac{1}{x}+\frac{1}{x}+\frac{1}{x}=\frac{3}{x},
$$

we get
$$
\frac{1}{x}<a=1<\frac{3}{x} \text {. }
$$

Thus,
$$
1<x<3 \text {. }
$$

Therefore, we must have
$$
x=2 \text {. }
$$

The given equation then becomes
$$
\begin{array}{c}
\frac{1}{2}+\frac{1}{y}+\frac{1}{z}=1 \\
\frac{1}{y}+\frac{1}{z}=\frac{1}{2} .
\end{array}
$$

Also, since
$$
\frac{1}{y}<\frac{1}{2}<\frac{2}{y} \\
2<y<4
$$

we must have
$$
y=3 \text {. }
$$

From the given equation, we then get
$$
z=6 \text {. }
$$

Therefore, the unique solution to this problem is: $(x, y, z)=(2,3,6)$.