Determine the smallest positive number $a$ such that the number of all integers belonging to $(a, 2016a]$ is $2016$.
1. We need to find the smallest positive number \( a \) such that the number of integers in the interval \( (a, 2016a] \) is 2016. This can be expressed mathematically as:
   \[
   \lfloor 2016a \rfloor - \lceil a \rceil + 1 = 2016
   \]

2. Since \( a \) is a positive number, we need to consider the range of \( a \). Given that \( \lceil a \rceil \) is the smallest integer greater than or equal to \( a \), and \( \lfloor 2016a \rfloor \) is the largest integer less than or equal to \( 2016a \), we can start by analyzing the possible values of \( \lceil a \rceil \).

3. Since \( \lfloor 2016a \rfloor - \lceil a \rceil + 1 = 2016 \), we can rearrange this equation to:
   \[
   \lfloor 2016a \rfloor = 2016 + \lceil a \rceil - 1
   \]

4. To minimize \( a \), we need to consider the smallest possible value for \( \lceil a \rceil \). Since \( a \) is positive and we are looking for the smallest \( a \), we start with \( \lceil a \rceil = 2 \) (since \( 1 < a < 2 \)).

5. Substituting \( \lceil a \rceil = 2 \) into the equation:
   \[
   \lfloor 2016a \rfloor = 2016 + 2 - 1 = 2017
   \]

6. Now, we need to find \( a \) such that \( \lfloor 2016a \rfloor = 2017 \). This implies:
   \[
   2016a \leq 2017 < 2016a + 1
   \]

7. Solving for \( a \):
   \[
   2016a \leq 2017 \implies a \leq \frac{2017}{2016}
   \]
   \[
   2017 < 2016a + 1 \implies 2016 < 2016a \implies a > \frac{2016}{2016} = 1
   \]

8. Combining these inequalities, we get:
   \[
   1 < a \leq \frac{2017}{2016}
   \]

9. The smallest \( a \) that satisfies this condition is:
   \[
   a = \frac{2017}{2016}
   \]

The final answer is \(\boxed{\frac{2017}{2016}}\).