Polynomial Ends here

Let $f(x) = x^{2018} + x^{2017} + \cdots + x^2 + x - 1345$ . Since $a_1, a_2, a_3, \cdots a_{2018}$ are roots of $f(x)$ , $\implies f(x) = \displaystyle \prod_{n=1}^ {2018} (x-a_n)$ . Now consider

$\begin{aligned} \sum_{n=1}^{2018} \frac 1{1-a_n} & = \frac {\sum_{k=1}^{2018}\prod_{n\ne k}^{2018} (1-a_n)}{\prod_{n=1}^{2018} (1-a_n)} \\ & = \frac {\sum_{k=1}^{2018} k\color{#3D99F6} c_k}{f(1)} & \small \color{#3D99F6} \text{where }c_k \text{ is the coefficient of }x^k \\ & = \frac {\sum_{k=1}^{2018} k}{2018-1345} & \small \color{#3D99F6} \text{in } f(x) \text{ and all }c_k = 1. \\ & = \frac {2018(2018+1)/2}{673} \\ & = \boxed{3027} \end{aligned}$

If we make the substitution $y = \tfrac{1}{1-x}$ , the equation becomes $\begin{aligned} \frac{1345}{x} & = \; 1+ x + x^2 + \cdots+ x^{2017} \\ 1345\frac{x-1}{x} & = \; x^{2018} - 1 \\ -1345\frac{1}{y-1} & = \; \left(\frac{y-1}{y}\right)^{2018} - 1 \\ y^{2018}(y-1) - (y-1)^{2019} - 1345y^{2018} & = \; 0 \end{aligned}$ and this last is a polynomial of degree $2018$ whose zeros are $\tfrac{1}{1-a_j}$ for $1 \le j \le 2018$ . Since $y^{2018}(y-1) - (y-1)^{2019} - 1345y^{2018} \; = \; \big(2019 - 1345 - 1\big)y^{2018} - \binom{2019}{2}y^{2017} + \cdots \; = \; 673y^{2018} - 2037171y^{2017} + \cdots$ we deduce that $\sum_{j=1}^{2018} \frac{1}{1-a_j} \; = \; \frac{2037171}{673} \; = \; \boxed{3027}$