jmc

Let $10a+b$ represent $n$, where $a$ and $b$ are the tens and units digits, respectively. Switching the digits and adding 3 results in $10b+a+3$, which we set equal to $2n$. $$\begin{array}{rl}2(10a+b)& =10b+a+3\Rightarrow \\ 20a+2b& =10b+a+3\Rightarrow \\ 19a& =8b+3\end{array}$$For the smallest $n$, we let the tens digit $a=1$. We have $19=8b+3$, which means $b=2$. So the smallest $n$ is $\boxed{12}$.