In fact, $N=(a+b+c)\cdot (\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\mathbb\in{Z}$ $\Rightarrow$ $abc\mid (a+b+c)\cdot (ab+bc+ca)$ $\Rightarrow$ $a\mid (a+b+c)\cdot (ab+bc+ca)$ $\Rightarrow$ $a\mid (b+c)\cdot bc$ use $gcd(a,b)=gcd(a,c)=1$ $\Rightarrow$ $a\mid b+c$ In the same way we can get $b\mid c+a$, $c\mid a+b$ Suppose instead $a\leq b\leq c$ use $gcd(a,b)=gcd(a,c)=gcd(b,c)=1$ easy to know (a,b,c)=(1,1,1),(1,1,2),(1,2,3) $\Rightarrow$ $${\boxed {N=9,10,11}}$$