Just for fun!-X

$\begin{aligned} \frac{2+4+6+\cdots+2014}{1+3+5+\cdots+2013}&=\frac{1+1+3+1+5+1+\cdots+2013+1}{1+3+5+\cdots+2013}\\&=\frac{1+3+5+\cdots+2013+\color{#3D99F6}{\boxed{1007}}}{1+3+5+\cdots+2013} \end{aligned}$

manual approach : if you follow the two series then its obvious that in every term there is a difference of 1. i.e .(2+4+6+..................)-(1+3+5+.............)=(2-1)+(4-3)+(6-5)+.............. so no. of terms of each series gives u the difference between this two series , which is the remainder . now, 1+(n-1)*2=2013 so , n =1007

Formula for Sum of first n even numbers can be found using AP i.e. n(n+1) and similarly for odd numbers we have (n+1)^2. Using these formulae expressions can be simplified.