67th Romanian Mathematical Olympiad

If one of the numbers is $0$, then all are $0$. If $abc\ne 0$, then the relation can be written $\frac{1}{bc}+\frac{1}{ac}+\frac{1}{ab}=1$. Since the relation is symmetric in $a,b,c$, we may assume that $a\le b\le c$, whence $ab\le ac\le bc$. If $ab>3$, then $\frac{1}{bc}+\frac{1}{ac}+\frac{1}{ab}<1$. If $ab=2$, then $a=1$ and $b=2$, whence $\frac{1}{2c}+\frac{1}{c}=\frac{1}{2}$, so $c=3$.

If $ab=3$, then $a=1$ and $b=3$, whence $\frac{1}{3c}+\frac{1}{c}=\frac{2}{3}$, so $c=2$, which contradicts $b<c$. Finally, the solutions are $(0,0,0)$ and all the permutations of $(1,2,3)$.