For each positive integer $n\ge 2$, define $k\left(n\right)$ to be the largest integer $m$ such that $\left(n!\right)^m$ divides $2016!$. What is the minimum possible value of $n+k\left(n\right)$?

[i]Proposed by Tristan Shin[/i]
1. To solve the problem, we need to find the largest integer \( m \) such that \((n!)^m\) divides \(2016!\). This can be determined by considering the prime factorization of \(2016!\) and \(n!\).

2. For a prime \( p \leq n \), the exponent of \( p \) in \( (n!)^m \) is given by:
   \[
   v_p((n!)^m) = m \sum_{k=1}^{\infty} \left\lfloor \frac{n}{p^k} \right\rfloor
   \]
   where \( v_p(x) \) denotes the exponent of \( p \) in the prime factorization of \( x \).

3. Similarly, the exponent of \( p \) in \( 2016! \) is:
   \[
   v_p(2016!) = \sum_{k=1}^{\infty} \left\lfloor \frac{2016}{p^k} \right\rfloor
   \]

4. For \((n!)^m\) to divide \(2016!\), we must have:
   \[
   m \sum_{k=1}^{\infty} \left\lfloor \frac{n}{p^k} \right\rfloor \leq \sum_{k=1}^{\infty} \left\lfloor \frac{2016}{p^k} \right\rfloor
   \]

5. Therefore, the largest \( m \) is given by:
   \[
   m = \min_{p \leq n} \left\lfloor \frac{\sum_{k=1}^{\infty} \left\lfloor \frac{2016}{p^k} \right\rfloor}{\sum_{k=1}^{\infty} \left\lfloor \frac{n}{p^k} \right\rfloor} \right\rfloor
   \]

6. To minimize \( n + k(n) \), we need to find the smallest \( n \) such that \( n + k(n) \) is minimized. We start by estimating \( n \) around \( \sqrt{2016} \approx 44.89 \).

7. We try different values of \( n \) and calculate \( k(n) \) for each:
   - For \( n = 41 \):
     \[
     k(41) = \min_{p \leq 41} \left\lfloor \frac{\sum_{k=1}^{\infty} \left\lfloor \frac{2016}{p^k} \right\rfloor}{\sum_{k=1}^{\infty} \left\lfloor \frac{41}{p^k} \right\rfloor} \right\rfloor
     \]
     After calculation, we find \( k(41) = 50 \), so \( n + k(n) = 41 + 50 = 91 \).

   - For \( n = 43 \):
     \[
     k(43) = \min_{p \leq 43} \left\lfloor \frac{\sum_{k=1}^{\infty} \left\lfloor \frac{2016}{p^k} \right\rfloor}{\sum_{k=1}^{\infty} \left\lfloor \frac{43}{p^k} \right\rfloor} \right\rfloor
     \]
     After calculation, we find \( k(43) = 47 \), so \( n + k(n) = 43 + 47 = 90 \).

   - For \( n = 47 \):
     \[
     k(47) = \min_{p \leq 47} \left\lfloor \frac{\sum_{k=1}^{\infty} \left\lfloor \frac{2016}{p^k} \right\rfloor}{\sum_{k=1}^{\infty} \left\lfloor \frac{47}{p^k} \right\rfloor} \right\rfloor
     \]
     After calculation, we find \( k(47) = 42 \), so \( n + k(n) = 47 + 42 = 89 \).

   - For \( n = 53 \):
     \[
     k(53) = \min_{p \leq 53} \left\lfloor \frac{\sum_{k=1}^{\infty} \left\lfloor \frac{2016}{p^k} \right\rfloor}{\sum_{k=1}^{\infty} \left\lfloor \frac{53}{p^k} \right\rfloor} \right\rfloor
     \]
     After calculation, we find \( k(53) = 39 \), so \( n + k(n) = 53 + 39 = 92 \).

8. From the above calculations, the minimum value of \( n + k(n) \) is \( 89 \) when \( n = 47 \).

The final answer is \(\boxed{89}\).