Answer: 63

Solution:

Write the integers from $0$ to $10^n - 1$ as $n$-digit strings with leading zeros. Each of the $n$ digit positions takes each digit value equally often, so each nonzero digit appears $n \cdot 10^{n-1}$ times. Adding the digit $1$ of $10^n$ itself, $$\begin{aligned} S_n &= 1 \\ &\quad {}+ n \\ &\quad {}\cdot 10^{n-1}\small\left(1 + \frac{1}{2} + \cdots + \frac{1}{9}\right) \\ &= 1 + \frac{7129}{2520}\, n \cdot 10^{n-1}. \end{aligned}$$

Since $\gcd(7129, 2520) = 1,$ the sum is an integer exactly when $2520 \mid n \cdot 10^{n-1}.$ Now $2520 = 2^3 \cdot 3^2 \cdot 5 \cdot 7,$ and for $n \ge 4$ the factor $10^{n-1}$ supplies $2^3 \cdot 5,$ leaving the condition $63 \mid n$ (a power of $10$ has no factors of $3$ or $7$). For $n = 1, 2, 3$ the products $1, 20, 300$ are not multiples of $2520.$

The smallest solution is therefore $n = 63.$