Inspired by Christian Daang

Relevant wiki: Power Mean Inequality (QAGH)

As $x_1, x_2,..., x_n$ are positive reals, we can be sure that:

${ n }^{ n-1 }({ x }_{ 1 }^{ n }+{ x }_{ 2 }^{ n }+...+{ x }_{ n }^{ n })\ge { ({ x }_{ 1 }+{ x }_{ 2 }+...+{ x }_{ n }) }^{ n }$

$<=> { n }^{ n-1 }\times { 2 }^{ 2016 }\ge (2n)^{ n }$

$<=> 2^{ 2016 }\ge { 2 }^{ n }n={ 2 }^{ n }\times { 2 }^{ \log _{ 2 }{ n } }={ 2 }^{ n+\log _{ 2 }{ n } }$

$<=> 2016\ge n+\log _{ 2 }{ n }$

It's easily seen that $1024 < n < 2048$ , so $11> \log _{ 2 }{ n } > 10$

$<=> n < 2006$

$<=>\max { (n)=2005 } (n \in N)$

Assume that $x_1=x_2=...=x_{2003}=2$ and $x_{2004} > x_{2005}$ , we have:

$\large \begin{cases} x_{ 2004 }+x_{ 2005 }=4\\{ x }_{ 2004 }^{ 2005 }+{ x }_{ 2005 }^{ 2005 }={ 2 }^{ 2005 }*45 \end{cases}$

Let $m, n$ be positive real numbers so that $m = 2*x_{2004}, n=2*x_{2005}$ , we have:

$\large \begin{cases} m + n = 2\\m^{2005} + n^{2005} = 45 \end{cases}$

At this point, you can either use intermediate value theorem or use your calculator ( $45$ is definitely small enough for most scientific calculators). If you use your calculator then you should find out (approximately) that $m = 1.0019001430699... , n = 0.9980998569301...$ , which also means that $x_{2004}=2.0038002861398... , x_{2005}=1.9961997138602...$ .

How to prove there exists a solution using IVT (Calvin): "We have $x_{2004} = y, x_{2005} = 4 - y$ , with $y \in [0, 2 ]$ . We want to show that $f(y) = y^{2005} + (4-y)^{2005} = 45 \times 2 ^ { 2005 }$ has a solution. This follows from IVT since $f(y)$ is a continuous function and $f( 2) = 2 \times 2 ^ { 2005 } < 45 \times 2 ^ { 2005 } < 4^{2005 } = f(0)$ "