Year $2018$ in an Exponent.

Let $f(x) = x^{2018} + x^{2017} + x^{2016} + \dots + x^2 + x - 1008.5$ . Since $a_1$ , $a_2$ , $\dots$ , $a_{2018}$ are the roots, $f(x) = (x - a_1)(x - a_2) \dotsm (x - a_{2018}).$ By logarithmic differentiation, $\frac{f'(x)}{f(x)} = \frac{1}{x - a_1} + \frac{1}{x - a_2} + \dots + \frac{1}{x - a_{2018}}.$ Thus, $\frac{1}{1 - a_1} + \frac{1}{1 - a_2} + \dots + \frac{1}{1 - a_{2018}} = \frac{f'(1)}{f(1)} = \frac{2018 + 2017 + \dots + 1}{2018 - 1008.5} = \frac{2018 \cdot 2019/2}{1009.5} = 2018.$

Relevant wiki: Vieta's Formula Problem Solving - Intermediate

$\begin{aligned} \sum_{n=1}^{2018} \frac 1{1-a_n} & = \frac {\displaystyle 2018 - 2017 \sum_{cyc} a_1 + 2016 \sum_{cyc} a_1a_2 - 2015 \sum_{cyc} a_1a_2a_3 + \cdots - \sum_{cyc} a_1a_2a_3 \cdots a_{2017}}{\displaystyle \prod_{n=1}^{2018}(1-a_n)} \end{aligned}$

Given that $a_1, a_2, a_3 \cdots a_{2018}$ are roots of $x^{2018} + 2^{2017} + x^{2016} + \cdots + x^2 + x - 1008.5 = 0$ , by Vieta's formula, we have $\displaystyle \sum_{cyc} a_1 a_2a_3\cdots a_n = (-1)^n$ and that $\displaystyle \prod_{n=1}^{2018}(x-a_n) = x^{2018} + 2^{2017} + x^{2016} + \cdots + x^2 + x - 1008.5$ $\implies \displaystyle \prod_{n=1}^{2018}(1-a_n) = 2018 - 1008.5 = 1009.5 = \dfrac {2019}2$ . Therefore,

$\begin{aligned} \sum_{n=1}^{2018} \frac 1{1-a_n} & = \frac {2018 + 2017 + 2016 + 2015 + \cdots + 1}{\displaystyle \prod_{n=1}^{2018}(1-a_n)} = \frac {\frac {2018\cdot 2019}2}{\frac {2019}2} = \boxed{2018} \end{aligned}$

Generalization:

For $\displaystyle \sum_{k=1}^n x^k = \frac {n-1}2$ , then $\displaystyle \sum_{k=1}^n \frac 1{1-a_k} = \frac {\sum_{k=1}^n k}{\prod_{k=1}^n (1-a_k)} = \frac {\frac {n(n+1)}2}{n - \frac {n-1}2} = n$

Note: The coefficients in the numerator come from $(-1)^k\dfrac {n\binom {n-1}k}{\binom nk} = (-1)^k \dfrac {n(n-1)!}{k!(n-k-1)!}\cdot \dfrac {k!(n-k)!}{n!} = (-1)^k(n-k)$ , for $k = 0, 1, 2, ... n$ .

Relevant wiki: Vieta's Formula Problem Solving - Intermediate

Let's begin by defining $b_n=\frac{1}{1-a_n}$ . Rearranging gives us $a_n=\frac{b_n -1}{b_n}$ . Since we know that $-1008.5+\sum_{k=0}^{2018}a^k_n =0$ for all $1\le n\le 2018$ , we can substitute $b_n$ in to get a new polynomial: $\sum_{k=0}^{2018} \Big(\frac{b_n -1}{b_n}\Big)^k -1008.5 =0 \implies \sum_{k=0}^{2018} (b_n)^{2018-k}(b_n-1)^k -1008.5 \ b_n^{2018} =0$

where we have multiplied both sides by $b_n^{2018}$ which is nonzero because $a_n \ne 1$ . This is true for all $1\le n \le 2018$ , so $b_n$ are the roots of the polynomial: $\sum_{k=0}^{2018}x^{2018}(x-1)^k - 1008.5 \ x^{2018}=0.$

By Vieta's it is enough to calculate the coefficients of $x^{2018}$ and $x^{2017}$ in the polynomial to compute the roots.

We see that the coefficients of $x^{2018}$ is $2018-1008.5=1009.5$ , and the coefficient of $x^{2017}= -1-2-\dots-2018=\frac{-(2018)(2019)}{2}.$

Thus, $\sum_{n=1}^{2018}\frac{1}{1-a_n}= \frac{2018\times2019}{2\times 1009.5}= \boxed{2018}$ .