$1 \cdot 59$ Determine all positive rational triples $(x, y, z)$ such that $x+y+z, \frac{1}{x}+\frac{1}{y}+\frac{1}{z}, x y z$ are all integers. (Here $x \leqslant y \leqslant z$.)
[Solution] Since
$$
x y+y z+z x=x y z\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right),
$$

thus $x y+y z+z x$ is also an integer.
Let
$$
\begin{aligned}
f(u) & =(u-x)(u-y)(u-z) \\
& =u^{3}-(x+y+z) u^{2}+(x y+y z+z x) u-x y z,
\end{aligned}
$$

then $f(u)$ is a cubic polynomial with integer coefficients. Since $x, y, z$ are three positive rational roots of $f(u)$, we let
$$
x=\frac{a}{b}, a, b \in \mathbb{N}, \text { and } (a, b)=1.
$$

Thus
$$
b^{2} f\left(\frac{a}{b}\right)=0
$$

which means
$$
\frac{a^{3}}{b}-(x+y+z) \cdot a^{2}+(x y+y z+z x) \cdot a b-x y z b^{2}=0.
$$

Except for the first term on the left side, all other terms are integers, so the first term $\frac{a^{3}}{b}$ is also an integer, which implies
$$
b=1, x=a, a \text { is a positive integer. }
$$

Similarly, $y, z$ are also positive integers.
If $x=1$, then $1+\frac{1}{y}+\frac{1}{z}$ is an integer, thus $\frac{1}{y}+\frac{1}{z}$ is an integer. If $y \geqslant 3$, then $z \geqslant y \geqslant 3$, so we have
$$
0<\frac{1}{y}+\frac{1}{z} \leqslant \frac{1}{3}+\frac{1}{3} \leqslant \frac{2}{3},
$$

which contradicts $\frac{1}{y}+\frac{1}{z}$ being an integer. Therefore, $y \leqslant 2$.
When $x=1, y=1$, only $z=1$ can make $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ an integer; when $x=1, y=2$, only $z=2$ can make $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ an integer. So when $x=1$, there are two solutions:
$$
\left\{\begin{array}{l}
x=1, \\
y=1, \\
z=1;
\end{array} \quad \left\{\begin{array}{l}
x=1 \\
y=2 \\
z=2
\end{array}\right.\right.
$$

If $x=2$, then when $y=2$, there is no positive integer $z \geqslant 2$ that can make $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ an integer; when $y=3$, only $z=6$ meets the condition; when $y=4$, only $z=4$ meets the condition; when $y \geqslant 5$, $z \geqslant 5$, so $\frac{1}{y}+\frac{1}{z} \leqslant \frac{2}{5}, \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \leqslant \frac{1}{2}+\frac{2}{5} < 1$, there is no solution. Therefore, when $x=2$, there are only two solutions:
$$
\left\{\begin{array}{l}
x=2, \\
y=3, \\
z=6;
\end{array} \quad \left\{\begin{array}{l}
x=2, \\
y=4, \\
z=4.
\end{array}\right.\right.
$$

If $x=3$, then $y \geqslant 3$. When $y=3$, obviously, there is only one solution:
$$
\left\{\begin{array}{l}
x=3 \\
y=3 \\
z=3
\end{array}\right.
$$

When $y \geqslant 4$, $z \geqslant 4$, so $\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \leqslant \frac{5}{6}<1$, no solution.
In summary, there are the five solutions mentioned above.