, 2015

All $n \in \mathbb{N}$ are representable except the powers of $2$. Set $f(a,b,c) = [a,b] + [b,c] + [c,a]$. Take an arbitrary $k \in \mathbb{N}$ and let $a = k$, $b = c = 1$ to obtain $f(k,1,1) = 2k + 1$. Hence all odd $n$, $n \ge 3$, are representable. If $n$ is representable then so is $2n$

because $[2u, 2v] = 2$ implies $f(2a, 2b, 2c) = 2f(a, b, c)$. So each $n \ge 3$ is representable if it has an odd divisor greater than $1$.
There remain the powers $2^k$ of $2$, with $k \ge 0$. We show that they are not representable. This is true for $k = 0, 1$ since clearly $f(a,b,c) \ge 3$ for all $a, b, c \in \mathbb{N}$. Suppose that $f(a,b,c) = 2^k$ with $k \ge 2$. Consider the least $k$ with this property. Observe that at least two numbers among $a, b, c$ are even. Otherwise $f(a,b,c)$ is odd while $2^k$ is even. If $a, b, c$ are all even, they can be divided by $2$ to yield $f\left(\frac{a}{2}, \frac{b}{2}, \frac{c}{2}\right) = 2^{k-1}$, which contradicts the minimality of $k$. Hence one may assume that $a, b$ are even and $c$ is odd. Here $[a,b] = 2\left[\frac{a}{2}, \frac{b}{2}\right]$. Note also that
$$[a,c] = 2\left[\frac{a}{2}, c\right] \text{ holds because } a \text{ is even and } c \text{ is odd.}$$
Analogously $[b,c] = 2\left[\frac{b}{2}, c\right]$. It follows that
$$f\left(\frac{a}{2}, \frac{b}{2}, c\right) = \frac{1}{2} f(a, b, c) = 2^{k-1}, \text{ contradicting the minimality of } k \text{ again.}$$