Take any $a$ and $b=c=1$ then $[a,b]+[a,c]+[b,c]=2a+1$ hence we find all odd numbers $>1$. Also we can multiply $a,b,c$ by a factor $2^k$ hence we find all numbers, except possibly the powers of 2. Now suppose $[a,b]+[a,c]+[b,c]=2^k (*)$ then $k\ge 2$. We may assume $k$ minimal, then $(a,b,c)=1$. Wlog $c$ odd. Also $a,b,c$ are not all odd, hence $a$ or $b$ is even. Write $2^s||a,2^t||b$. If $s<t$ then the LHS in (*) contains exactly $s$ factors 2, hence $s=k$, impossible. In the same way $t<s$ is impossible hence $s=t\ge 1$. Then $2^s||[a,b],[a,c],[b,c]$ hence $2^s||[a,b]+[a,c]+[b,c]$ since the sum of three odd integers is odd. Hence $s=k$, contradiction. Thank you nosquares for correcting my earlier mistake.

alexheinis wrote: Take any $a$ and $b=c=1$ then $[a,b]+[a,c]+[b,c]=2a+1$ hence we find all odd numbers. Also we can multiply $a,b,c$ by a factor $2^k$ hence we find all numbers. Ahhh, not exactly. You can't get $1$ and $2$, so it is not obviously why you can get $2^k$... (If I'm not mistaken, $n=2^k$ doesn't work)