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$\Large\displaystyle\sum_{z=0}^{n}z=\dfrac{n(n+1)}{2}$ $\Large \therefore \color{#EC7300}{n}(\color{darkviolet}{n+1})=4032=\color{#EC7300}{63}\times \color{darkviolet}{64}$ Hence $\large\boxed{\color{#D61F06}{n=63}}$

We know that sum of $n$ natural numbers is $=\displaystyle\sum_{z=0}^{n}z=\dfrac{n(n+1)}{2}$ .

In this case,

$\displaystyle\sum_{z=0}^{n}z=\dfrac{n(n+1)}{2}=2016\\\implies n^2+n-4032=0\\ n=\dfrac{-1\pm\sqrt{1-(4\times 1\times 4032}}{2}=\dfrac{-1\pm127}2$

Therefore, $\boxed{n=63}$ or $n=-64$ , as $n$ can't be negative, the positive value is correct.