For a positive integer $n$, let $S(n)$ be the sum of its decimal digits. Determine the smallest positive integer $n$ for which $4 \cdot S(n)=3 \cdot S(2n)$.
Assume we carry $C$ times in the addition problem $n+n$. Then $S(2n)=2S(n)-9C$. So we have $$\frac{2S(n)-9C}{S(n)}=\frac{4}{3}\iff \frac{9C}{S(n)}=\frac{2}{3}$$So we must have that $S(n)$ is at least $27$. If $S(n)=27$ we must have $C=2$ so we have exactly two digits of $n$ at least $5$. The smallest such $n$ is $\boxed{14499}$, which works. The cases $S(n)>27$ can be readily checked.