Let's determine all triples $(p ; q ; r)$ of positive rational numbers for which $p+q+r$, $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}$, and $p q r$ are all integers.
Solution. Let the three numbers in question be denoted by $a, b, c$. Since $b c=p q+q r+r s$, according to the relationships between the roots and coefficients (Viète's formulas), $p, q, r$ are precisely the three roots of the polynomial $(x-p)(x-q)(x-r)=x^{3}-a x^{2}+b c x-c$. According to the condition, $a, b, c$ are integers, so $b c$ is also an integer, which means this polynomial has integer coefficients. Its leading coefficient is 1, so every rational root is an integer. If $x_{0}=\frac{u}{v}$ is a root, where $u$ and $v$ are relatively prime integers, then after substituting $x_{0}$ and multiplying by $v^{3}$: $0 \cdot v^{3}=u^{3}-a u^{2} v+b c u v^{2}-c v^{3}=u^{3}-v\left(a u^{2}-b c u v+c v^{2}\right)$. Since $(u ; v)=1$, $u^{3}$ and $v$ are also relatively prime, but the above equality implies that $v$ divides $u^{3}$. This is only possible if $v= \pm 1$, which means $x_{0}=\frac{u}{v}$ is an integer.

Therefore, the sought $p, q, r$ numbers are positive integers whose sum, product, and reciprocal sum are all integers. The sum and product are always integers, so we need to examine when $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}$ is also an integer. Let $p \leq q \leq r$. If $3 \leq p$, then

$$
0<\frac{1}{p}+\frac{1}{q}+\frac{1}{r} \leq \frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1
$$

thus the reciprocal sum is an integer only if equality holds, i.e., $p=q=r=3$. If $p=2$, then

$$
0<\frac{1}{p}+\frac{1}{q}+\frac{1}{r} \leq \frac{3}{2}, \quad \text { therefore } \quad \frac{1}{p}+\frac{1}{q}+\frac{1}{r}=1
$$

from which

$$
\frac{1}{q}+\frac{1}{r}=\frac{1}{2}, \quad(q-2)(r-2)=4
$$

so $q=r=4$, or $q=3, r=6$. Finally, if $p=1$, then

$$
0<\frac{1}{q}+\frac{1}{r} \leq 2, \quad \text { so } \quad \frac{1}{q}+\frac{1}{r}
$$

has a value of 1 or 2, from which $q=r=2$, or $q=r=1$ respectively.

Therefore, all the suitable $(p ; q ; r)$ triples, disregarding the order of the numbers, are the following: $(1 ; 1 ; 1)$, $(1 ; 2 ; 2),(2 ; 4 ; 4),(2 ; 3 ; 6)$ and $(3 ; 3 ; 3)$.