Solution:
We will first calculate $S_{999}$, then $S_{1999} - S_{999}$, and then $S_{2016} - S_{1999}$.

Writing the integers from $1$ to $999$ as $001$ to $999$, adding eventually also $000$ (since $0$ digits actually do not matter), each digit appears exactly $100$ times in each position (as unit, ten, or hundred). Hence
$$S_{999} = 300 \cdot \left( \frac{1}{1} + \frac{1}{2} + \cdots + \frac{1}{9} \right)$$
For the numbers in the interval $1000 \rightarrow 1999$, compared to $0 \rightarrow 999$, there are precisely $1000$ more digits $1$. We get
$$S_{1999} - S_{999} = 1000 + S_{999} \Longrightarrow S_{1999} = 1000 + 600 \cdot \left( \frac{1}{1} + \frac{1}{2} + \cdots + \frac{1}{9} \right)$$
Finally, in the interval $2000 \rightarrow 2016$, the digit $1$ appears $9$ times as unit and $19$ times as a ten, the digit $2$ twice as a unit and $17$ times as a thousand, the digits $3, 4, 5$, and $6$ each appear exactly twice as units, and the digits $7, 8, 9$ each appear exactly once as a unit. Hence
$$S_{2016} - S_{1999} = 9 \cdot 1 + 19 \cdot \frac{1}{2} + 2 \cdot \left( \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} \right) + 1 \cdot \left( \frac{1}{7} + \frac{1}{8} + \frac{1}{9} \right)$$
In the end, we get
$$\begin{aligned} S_{2016} &= 1609 \cdot 1 + 619 \cdot \frac{1}{2} + 602 \cdot \left( \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} \right) + 601 \cdot \left( \frac{1}{7} + \frac{1}{8} + \frac{1}{9} \right) \\ &= m + \frac{1}{2} + \frac{2}{3} + \frac{2}{4} + \frac{2}{5} + \frac{2}{6} + \frac{6}{7} + \frac{1}{8} + \frac{7}{9} = n + \frac{p}{2^{3} \cdot 3^{2} \cdot 5 \cdot 7} \end{aligned}$$
where $m, n$, and $p$ are positive integers, $p$ coprime to $2^{3} \cdot 3^{2} \cdot 5 \cdot 7$. Then $k! \cdot S_{2016}$ is an integer precisely when $k!$ is a multiple of $2^{3} \cdot 3^{2} \cdot 5 \cdot 7$. Since $7 \mid k!$, it follows that $k \geq 7$. Also, $7! = 2^{4} \cdot 3^{2} \cdot 5 \cdot 7$, implying that the least $k$ satisfying $k! \cdot S_{2016} \in \mathbb{Z}$ is $k = 7$.