Let $S_{n}$ be the sum of reciprocal values of non-zero digits of all positive integers up to (and including) $n$. For instance, $S_{13}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{1}+\frac{1}{1}+\frac{1}{1}+\frac{1}{1}+\frac{1}{2}+\frac{1}{1}+\frac{1}{3}$. Find the least positive integer $k$ making the number $k!\cdot S_{2016}$ an integer.
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 twice as unit and seven 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, and 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 \( k = 7 \).