I am looking for pairwise coprime natural numbers $a$ $b$ $c$ for which $n = \dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}}$ is also a natural number.
I can find examples (note $a$ $b$ $c$ must differ) for which one of their three gcds is $1$ and the two other are prime numbers.
Maybe such triplets do not exist?
Multiply through by $abc$ in the numerator and denominator, this shows that we must have $ab+ac+bc$ divides $abc$. Suppose $p$ divides $ab+ac+bc$, then $p$ must divide $abc$. By Euclid's lemma and without loss of generality we may assume $p$ divides $a$. But if $p$ divides $a$ and $ab+ac+bc$, then it must also divide $bc$, which is a contradiction since they're relatively prime to $a$.