2017 is 2 years old

$\begin{aligned} Q & = \frac {\frac {2016}1+\frac {2015}2 + \frac {2014}3+\cdots + \frac 1{2016}}{\frac 12 + \frac 13+\frac 14 + \cdots + \frac 1{2017}} \\ & = \frac {\frac {2017-1}1+\frac {2017-2}2 + \frac {2017-3}3+\cdots + \frac {2017-2016}{2016}}{\frac 12 + \frac 13+\frac 14 + \cdots + \frac 1{2017}} \\ & = \frac {2017 \left(\frac 11+\frac 12 + \frac 13+\cdots + \frac 1{2016}\right) - 2016}{\frac 12 + \frac 13+\frac 14 + \cdots + \frac 1{2017}} \\ & = \frac {2017 \left(\frac 12 + \frac 13+\cdots + \frac 1{2016}\right) + 1}{\frac 12 + \frac 13+\frac 14 + \cdots + \frac 1{2017}} \\ & = \frac {2017 \left(\frac 12 + \frac 13+\cdots + \frac 1{2016} + \frac 1{2017} \right)}{\frac 12 + \frac 13+\frac 14 + \cdots + \frac 1{2017}} \\ & = \boxed{2017} \end{aligned}$

Notice that $\frac{2016}{1} + \frac{2015}{2} + \dots + \frac{1}{2016} = 2017 \left(1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{2016} \right) - 2016 = 2017 \left( \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots + \frac{1}{2017} \right)$ Therefore, the conclusion holds.

<?php
$n=0;
$d=0;
for($i=1;$i<2017;$i++){
$n += (2017-$i)/$i;
$d += 1/($i+1);
}
echo $n/$d; //prints 2017
?>