Seems Like a Popular Problem!

Relevant wiki: Telescoping Series - Product

First, let $x = 2017$

We know that $1 - \frac{1}{2^2} = (1 + \frac{1}{2} )( 1 - \frac{1}{2})$ .

Do the same thing for the other terms, we have

$\left(1 - \frac{1}{2} \right)\left(1 + \frac{1}{2} \right)\left(1- \frac{1}{3} \right)\left( 1+ \frac{1}{3}\right)\cdots \left(1 + \frac{1}{x}\right)$

Simplify, we have

$\left( \frac{1}{2}\right)\left( \frac{3}{2}\right)\left( \frac{2}{3}\right)\left( \frac{4}{3}\right)\left( \frac{3}{4}\right)\cdots \left( \frac{x+1}{x} \right) = \left(\frac{1}{2}\right)\left(1 + \frac{1}{x}\right) = \frac{1}{2} + \frac{1}{2x}$

Put $x=2017$ , we have

$\frac{1}{2} + \frac{1}{4034} = \frac{1}{n} + \frac{1}{m}$

$n =2$ and $m = 4034$

Hence, $2017n + 2m = 4034 + 8068 = \boxed{12102}$ .

Relevant wiki: Telescoping Series - Product

$\begin{aligned} P & = \left(1-\frac 1{2^2}\right) \left(1-\frac 1{3^2}\right) \left(1-\frac 1{4^2}\right) ... \left(1-\frac 1{2017^2}\right) \\ & = \prod_{n=2}^{2017} \left(1-\frac 1{n^2}\right) \\ & = \prod_{n=2}^{2017} \frac {n^2-1}{n^2} \\ & = \prod_{n=2}^{2017} \frac {(n-1){\color{#3D99F6}(n+1)}}{n^2} \\ & = \frac {2016! \cdot {\color{#3D99F6}2018!}}{(2017!)^2\cdot {\color{#3D99F6}2}} = \frac {2018}{2017\cdot 2} = \frac {1009}{2017} = \frac 12+\frac 1{4034} \end{aligned}$

$\implies 2017n+2m = 2017 \cdot 2 + 2 \cdot 4034 = \boxed{12102}$