Question 29
Find all $n \in \mathbb{N}$ for which there exist nonnegative integers $a_{1}, a_{2}, \ldots, a_{n}$ such that
$$
\frac{1}{2^{a_{1}}}+\frac{1}{2^{a_{2}}}+\cdots+\frac{1}{2^{a_{n}}}=\frac{1}{3^{a_{1}}}+\frac{2}{3^{a_{2}}}+\cdots+\frac{n}{3^{a_{n}}}=1 .
$$
$$
\frac{1}{2^{a_{1}}}+\frac{1}{2^{a_{2}}}+\cdots+\frac{1}{2^{a_{n}}}=\frac{1}{3^{a_{1}}}+\frac{2}{3^{a_{2}}}+\cdots+\frac{n}{3^{a_{n}}}=1 .
$$