4-193 Let $x, y, z$ be three distinct natural numbers, arranged in ascending order, and the sum of their reciprocals is also an integer. Find $x, y, z$.
[Solution] This problem is to find positive integers $x, y, z, a$ such that $x<y<z$,
and $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=a$.
Since for any $x, y, z$, we have
$$
a=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}<\frac{1}{1}+\frac{1}{2}+\frac{1}{2}=2,
$$

it follows that $a=1$. Also, from
$$
\frac{1}{x}<a=1<\frac{1}{x}+\frac{1}{x}+\frac{1}{x}=\frac{3}{x},
$$

we get $1<x<3$, hence $x=2$.
At this point, equation (1) becomes $\frac{1}{y}+\frac{1}{z}=\frac{1}{2}$,
which implies $\frac{1}{y}<\frac{1}{2}<\frac{2}{y}$,
$$
2<y<4 \text{, therefore } y=3 \text{. }
$$

Finally, from the equation
$$
\frac{1}{2}+\frac{1}{3}+\frac{1}{z}=1
$$

we can solve for $z=6$.
The only solution to this problem is: $a=1, x=2, y=3, z=6$.