We had this in the QEDMO (an olympiad organized by Darij and me in the QED association), so I just copy the solution I proposed there: We use induction: For $n=1=2^0 3^0$, it's true. If $n>1$ is even, then $n=2m$ with positive integer $m$. Now $m$ is by induction hypothesis of the desired form, $m=2^{a_1} 3^{b_1} + 2^{a_2} 3^{b_2} + ... + 2^{a_k} 3^{b_k}$. But then $n=2^{a_1+1} 3^{b_1} + 2^{a_2+1} 3^{b_2} + ... + 2^{a_k+1} 3^{b_k}$ is also of the desired form. If $n>1$ is odd, then let $n=2^0 3^p$ if $n$ is a power of $3$ and we are done then. Otherwise $n-3^p$ is even and $m=\frac{n-3^p}{2}$ is of desired type, $m=2^{a_1} 3^{b_1} + 2^{a_2} 3^{b_2} + ... + 2^{a_k} 3^{b_k}$. This gives $n=2^{a_1+1} 3^{b_1} + 2^{a_2+1} 3^{b_2} + ... + 2^{a_k+1} 3^{b_k} + 2^0 3^p$. This fulfills all requirements, because $0< a_i+1$ and $p>b_i$ fr alle $i$, such that the new summand does neither divide one of the old ones nor is divided by one of them.