jmc

Let $K={\sum }_{i=1}^9\frac{1}{i}$. Examining the terms in $S_1$, we see that $S_1=K+1$ since each digit $n$ appears once and 1 appears an extra time. Now consider writing out $S_2$. Each term of $K$ will appear 10 times in the units place and 10 times in the tens place (plus one extra 1 will appear), so $S_2=20K+1$. In general, we will have that $S_n=(n10^{n-1})K+1$ because each digit will appear $10^{n-1}$ times in each place in the numbers $1,2,\dots ,10^n-1$, and there are $n$ total places. The denominator of $K$ is $D=2^3\cdot 3^2\cdot 5\cdot 7$. For $S_n$ to be an integer, $n10^{n-1}$ must be divisible by $D$. Since $10^{n-1}$ only contains the factors $2$ and $5$ (but will contain enough of them when $n\ge 3$), we must choose $n$ to be divisible by $3^2\cdot 7$. Since we're looking for the smallest such $n$, the answer is $\boxed{63}$.