$ \Rightarrow mln|(m+n+l)(mn+ml+nl)$ But $ \gcd(m,mn+ml+nl)=\gcd(m,nl)=1$ Samilar :$ \gcd(n,nl+mn+lm)=\gcd(l,mn+nl+ml)=1$ Imply that $ \gcd(mnl,mn+ml+nl)=1$ So $ mnl|m+n+l$ Suppose $ m\geq n\geq l$ $ \frac{1}{ml}+\frac{1}{nl}+\frac{1}{mn}\in N$ But $ \frac{1}{mn}+\frac{1}{nl}+\frac{1}{ml}\leq 3$ so we only to consider three case. Case 1 $ m+n+l=mnl$ $ (m,n,l)=(3,2,1)$ Case 2 $ m+n+l=2mnl$ $ 3m\geq 2ln$ so $ nl=1\Rightarrow n=l=1,m=2$ Case 3$ m+n+l=3mnl$ so $ m=n=l=1$ So we has solution : $ (m,n,l)=(1,1,1),(3,2,1),(2,1,1)$ and cyclic.

A slightly different approach: Assume: $ n\geq m\geq l$ Case 1) l = m. Then l = m = 1. (2 + n)*(2 + 1/n) must be an integer; 2/n must be an integer; n = 1 or n = 2. This corresponds with the triples (l, m, n) = (1, 1, 1) and (l, m, n) = (1, 1, 2) Case 2) n > m > l. We have: (l + m + n)*(1/l + 1/m + 1/n) = (l + m + n)*(mn + ln + lm)/lmn; m | (l + n) and n | (l + m). Since l + m < n + n = 2n, we must have: l + m = n. Let m = (p/q)*n with (p, q) = 1. Now q < 2p, because: n = l + m < m + m = 2m. Now: m | (l + n) (p/q)*n | (n - (p/q)*n + n) pn | (2qn - pn) p | (2q - p) p | 2q p | 2 p = 1 is impossible since q < 2p, so p = 2, which implies q = 3 and we have the third triple: (l, m, n) = (1, 2, 3)

Answer: (1,1,1), (1,1,2), (1,2,3) and all permutations. Bringing the second term to a common denominator, we get $ lmn \mid (l+m+n)(lm+mn+nl)$ and, in particular, $ l \mid (l+m+n)(lm+mn+nl) \Rightarrow l \mid (m + n)mn \Rightarrow l\mid m + n$; likewise $ m \mid n + l$ and $ n \mid l + m$. Now assume WLOG that $ l\geq m\geq n$. Then $ (m+n)/l = 1$ or $ 2$; if the latter we have $ l = m = n = 1$. In the case $ m+n = l$, we get $ m \geq l/2 \Rightarrow l/m \leq 2$ and $ (n + l)/m \leq 3$. Moreover $ l \geq m$ so this ratio cannot be $ 1$; hence $ n + l = 2m$ giving the $ (1,2,3)$ solution or $ n + l = 3m$ giving the $ (1,1,2)$ solution.