Classical Inequalities addicts can do this. Part 1

Relevant wiki: Cauchy-Schwarz Inequality

Let's generalize this.

Let $x$ and $y$ denote the values

$\begin{aligned} x& =& x_1 + x_2 + \cdots + x_n \\ y& =& x_1 ^4 + x_2^4 + \cdots + x_n ^4 \end{aligned}$

By Cauchy-Schwarz inequality (CS), for all sequences of real numbers $a_i$ and $b_i$ , we have $\left(\displaystyle \sum_{i=1}^n a_i^2\right)\left( \displaystyle \sum_{i=1}^n b_i^2\right)\ge \left( \displaystyle \sum_{i=1}^n a_ib_i\right)^2$ .

Now suppose $a_i = x_i$ for $i = 1,2,3, \ldots, n$ and $b_1 = b_2 = \cdots = b_n = 1$ , then we have

(x_1^2 + x_2 ^2 + \cdots + x_n ^2 ) \left( \underbrace{1 + 1 + 1 + \cdots + 1}_{n \text{ number of 1's}} \right) \geq (x_1 + x_2 + \cdots + x_n)^2 .

Or equivalently, \color{#3D99F6}{x_1^2 + x_2 ^2 + \cdots + x_n ^2 \geq \dfrac{x^2}n} .

Now suppose we apply CS again, but this time, suppose $a_i = x_i^2$ for $i = 1,2,3, \ldots, n$ and $b_1 = b_2 = \cdots = b_n = 1$ , then we have

(x_1^4 + x_2 ^4 + \cdots + x_n ^4 ) \left( \underbrace{1 + 1 + 1 + \cdots + 1}_{n \text{ number of 1's}} \right) \geq (x_1^2 + x_2 ^2+ \cdots + x_n^2 )^2 .

Or equivalently, $\color{#3D99F6}{y \geq \dfrac{ (x_1^2 + x_2 ^2+ \cdots + x_n^2 )^2 } {n} }$ .

Combining these 2 inequalities (highlighted in blue) gives

$y \geq \dfrac{x^4}{n^3}.$

In this case, $x = 2016, y = 2016\times 512$ . Upon simplifying, we get $n\geq \boxed{252}$ , with $x_1 = x_2 = \cdots = x_{252} = \dfrac{2016}{252} = 8$ .