Note that \[\Big(\frac{1}{2}\Big)\Big(\frac{2}{3}\Big)=\Big(\frac{4}{8}\Big)\Big(\frac{2}{3}\Big)=\Big(\frac{3+1}{2^3}\Big)\Big(\frac{1+1}{3^1}\Big)=\frac{(3+1)(1+1)}{2^3\cdot3^1}\Rightarrow n=2^3\cdot3^1=\boxed{24}.\]Hence, $n=24$ works too.

First, we rearrange the above equation to make use of the multiplicative property. \[\frac{\tau(n)}{n}=\frac{(e_1+1)(e_2+1)\cdots(e_k+1)}{p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}}=\prod_{i=1}^k \frac{(e_i+1)}{p_i^{e_i}}=\frac{1}{3}.\]Note that exponential functions grow faster than linear functions. Let's analyse small values of $\frac{e+1}{p^e}$. \begin{tabular}{|c|c|c|c|c|} \hline & $p=2$ & $p=3$ & $p=5$ & $p=7$ \\ \hline $k=1$ & $\color{green}{\frac{1+1}{2^1}=1}$ & $\color{green}{\frac{1+1}{3^1}=\frac{2}{3}}$ & $\color{green}{\frac{1+1}{5^1}=\frac{2}{5}}$ & $\color{red}{\frac{1+1}{7^1}=\frac{2}{7}}$ \\ \hline $k=2$ & $\color{green}{\frac{2+1}{2^2}=\frac{3}{4}}$ & $\color{green}{\frac{2+1}{3^2}=\frac{1}{3}}$ & $\color{red}{\frac{2+1}{5^2}=\frac{3}{25}}$ & $\color{red}{\vdots}$ \\ \hline $k=3$ & $\color{green}{\frac{3+1}{2^3}=\frac{1}{2}}$ & $\color{red}{\frac{3+1}{3^3}=\frac{4}{27}}$ & $\color{red}{\vdots}$ & \\ \hline $k=4$ & $\color{red}{\frac{4+1}{2^4}=\frac{5}{16}}$ & $\color{red}{\vdots}$ & & \\ \hline \end{tabular}All the red boxes have values of $\frac{e+1}{p^e}$ greater than $\frac{1}{3}$, so we can disregard these boxes. This reduces our search to just the green boxes above. Inspection gives us that the only solutions are: \[\frac{1}{3}=\frac{3}{9}=\frac{(2+1)}{3^2}\Rightarrow n=3^2=\boxed{9}.\]\[(1)\Big(\frac{1}{3}\Big)=\Big(\frac{2}{2}\Big)\Big(\frac{3}{9}\Big)=\Big(\frac{1+1}{2^1}\Big)\Big(\frac{2+1}{3^2}\Big)=\frac{(1+1)(2+1)}{2^1\cdot3^2}\Rightarrow n=2^1\cdot3^2=\boxed{18}.\]\[\Big(\frac{1}{2}\Big)\Big(\frac{2}{3}\Big)=\Big(\frac{4}{8}\Big)\Big(\frac{2}{3}\Big)=\Big(\frac{3+1}{2^3}\Big)\Big(\frac{1+1}{3^1}\Big)=\frac{(3+1)(1+1)}{2^3\cdot3^1}\Rightarrow n=2^3\cdot3^1=\boxed{24}.\]