Let's see how good you are at polynomial transformation

Let $S_{n}$ denote the sum of the $n$ th powers of the roots of the given polynomial. From Newton's sums, we have $S_{1} = 0$ and $S_{2} = 0$ . This means that the sum of any $2013$ roots of this polynomial equals the negative of the last root: $a_{1} + a_{2} + ... + a_{n-1} + a_{n+1} + ... + a_{2014} = -a_{n}$ . Same thing for the squares of the roots: $a_{1}^{2} + a_{2}^{2} + ... + a_{n-1}^{2} + a_{n+1}^{2} + ... + a_{2014}^{2} = -a_{n}^{2}$ . Thus,

$\frac{r_{2}+r_{3}+\cdots+r_{2014}-r_{1}}{r_{2}^{2}+r_{3}^{2}+\cdots+r_{2014}^{2}-r_{1}^{2}} = \frac{-r_{1} -r_{1}}{-r_{1}^{2}-r_{1}^{2}} = \frac{-2r_{1}}{-2r_{1}^{2}} = \frac{1}{r_{1}}.$

Similarly, the other expressions evaluate as $\frac{1}{r_{2}}, \frac{1}{r_{3}}, ..., \frac{1}{r_{2014}}.$ The polynomial with these roots is $2014x^{2014} + x^{2000} + x^{1994} + 1,$ so $a+b+c+d+e=\boxed{8023}.$