Find all $x \in  R$ such that  $$ x - \left[ \frac{x}{2016} \right]= 2016$$, where $[k]$ represents the largest smallest integer or equal to $k$.
1. We start with the given equation:
   \[
   x - \left\lfloor \frac{x}{2016} \right\rfloor = 2016
   \]
   where $\left\lfloor k \right\rfloor$ represents the floor function, which gives the largest integer less than or equal to $k$.

2. Let $n = \left\lfloor \frac{x}{2016} \right\rfloor$. Then the equation becomes:
   \[
   x - n = 2016 \implies x = n + 2016
   \]

3. Since $n = \left\lfloor \frac{x}{2016} \right\rfloor$, we have:
   \[
   n \leq \frac{x}{2016} < n + 1
   \]

4. Substituting $x = n + 2016$ into the inequality:
   \[
   n \leq \frac{n + 2016}{2016} < n + 1
   \]

5. Simplifying the left part of the inequality:
   \[
   n \leq \frac{n + 2016}{2016} \implies n \leq \frac{n}{2016} + 1 \implies 2016n \leq n + 2016 \implies 2015n \leq 2016 \implies n \leq \frac{2016}{2015}
   \]
   Since $n$ is an integer, $n \leq 1$.

6. Simplifying the right part of the inequality:
   \[
   \frac{n + 2016}{2016} < n + 1 \implies \frac{n}{2016} + 1 < n + 1 \implies \frac{n}{2016} < n \implies n < 2016
   \]
   Since $n$ is an integer, $n < 2016$.

7. Combining the results from steps 5 and 6, we get:
   \[
   0 \leq n \leq 1
   \]

8. Since $n$ must be an integer, the only possible value for $n$ is 1.

9. Substituting $n = 1$ back into the equation $x = n + 2016$:
   \[
   x = 1 + 2016 = 2017
   \]

10. Verifying the solution:
    \[
    \left\lfloor \frac{2017}{2016} \right\rfloor = \left\lfloor 1.000495 \right\rfloor = 1
    \]
    \[
    2017 - 1 = 2016
    \]
    The solution satisfies the original equation.

The final answer is $\boxed{2017}$