find all positive integer $n$, satisfying
$$\frac{n(n+2016)(n+2\cdot 2016)(n+3\cdot 2016) . . . (n+2015\cdot 2016)}{1\cdot 2 \cdot 3 \cdot . . . . . \cdot 2016}$$ is positive integer.
To solve the problem, we need to find all positive integers \( n \) such that the expression 

\[
\frac{n(n+2016)(n+2 \cdot 2016)(n+3 \cdot 2016) \cdots (n+2015 \cdot 2016)}{1 \cdot 2 \cdot 3 \cdots 2016}
\]

is a positive integer. Let's denote the numerator and the denominator separately for clarity:

Numerator: \( n(n+2016)(n+2 \cdot 2016)(n+3 \cdot 2016) \cdots (n+2015 \cdot 2016) \)

Denominator: \( 1 \cdot 2 \cdot 3 \cdots 2016 = 2016! \)

We need the fraction \(\frac{n(n+2016)(n+2 \cdot 2016) \cdots (n+2015 \cdot 2016)}{2016!}\) to be an integer. This means that the numerator must be divisible by \(2016!\).

1. **Prime Factorization of 2016:**
   \[
   2016 = 2^5 \cdot 3^2 \cdot 7
   \]

2. **Divisibility Condition:**
   For the fraction to be an integer, the numerator must contain at least the same number of each prime factor as the denominator. Therefore, we need to ensure that the numerator has at least 5 factors of 2, 2 factors of 3, and 1 factor of 7.

3. **Considering Prime Factors:**
   - For \( p = 2 \):
     The sequence \( n, n+2016, n+2 \cdot 2016, \ldots, n+2015 \cdot 2016 \) must collectively contain at least 5 factors of 2.
   - For \( p = 3 \):
     The sequence must collectively contain at least 2 factors of 3.
   - For \( p = 7 \):
     The sequence must collectively contain at least 1 factor of 7.

4. **Ensuring Divisibility:**
   To ensure that the sequence contains the required number of factors of each prime, \( n \) must be divisible by the least common multiple (LCM) of the prime factors' requirements. The LCM of 2, 3, and 7 is 42.

5. **Conclusion:**
   Therefore, \( n \) must be divisible by 42. Any positive integer \( n \) that is a multiple of 42 will satisfy the condition.

\[
\boxed{n = 42k \text{ for any positive integer } k}
\]