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.
1. **Understanding the Problem:**
   We need to find the least positive integer \( k \) such that \( k! \cdot S_{2016} \) is an integer, where \( S_n \) is the sum of the reciprocals of the non-zero digits of all positive integers up to \( n \).

2. **Analyzing \( S_{2016} \):**
   To find \( S_{2016} \), we need to sum the reciprocals of all non-zero digits of numbers from 1 to 2016. For example, for \( n = 13 \), we have:
   \[
   S_{13} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{1} + \frac{1}{1} + \frac{1}{1} + \frac{1}{1} + \frac{1}{2} + \frac{1}{1} + \frac{1}{3}
   \]

3. **Generalizing the Sum:**
   We need to consider the frequency of each digit from 1 to 9 in the numbers from 1 to 2016. Each digit appears in each decimal place (units, tens, hundreds, etc.) with a certain frequency.

4. **Frequency Calculation:**
   - Units place: Each digit from 1 to 9 appears 201 times (since there are 2016 numbers, and each digit appears once in every 10 numbers).
   - Tens place: Each digit from 1 to 9 appears 201 times.
   - Hundreds place: Each digit from 1 to 9 appears 201 times.
   - Thousands place: Only the digit 1 appears 1000 times (from 1000 to 1999).

5. **Summing the Reciprocals:**
   \[
   S_{2016} = 201 \left( \sum_{d=1}^{9} \frac{1}{d} \right) + 201 \left( \sum_{d=1}^{9} \frac{1}{d} \right) + 201 \left( \sum_{d=1}^{9} \frac{1}{d} \right) + 1000 \cdot \frac{1}{1}
   \]
   Simplifying, we get:
   \[
   S_{2016} = 603 \left( \sum_{d=1}^{9} \frac{1}{d} \right) + 1000
   \]

6. **Harmonic Sum:**
   The sum of the reciprocals of the digits from 1 to 9 is:
   \[
   \sum_{d=1}^{9} \frac{1}{d} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9}
   \]

7. **Finding the Least \( k \):**
   To make \( k! \cdot S_{2016} \) an integer, \( k! \) must clear the denominators of the harmonic sum. The denominators are 2, 3, 4, 5, 6, 7, 8, and 9. The least common multiple (LCM) of these numbers is:
   \[
   \text{LCM}(2, 3, 4, 5, 6, 7, 8, 9) = 2520
   \]
   The smallest \( k \) such that \( k! \) is divisible by 2520 is \( k = 7 \) because \( 7! = 5040 \).

8. **Conclusion:**
   Therefore, the least positive integer \( k \) making \( k! \cdot S_{2016} \) an integer is \( k = 7 \).

The final answer is \( \boxed{ 7 } \)