jmc

Let $b_n=\frac{1}{1-a_n}.$ Solving for $a_n,$ we find $$a_n=\frac{b_n-1}{b_n}.$$Substituting, we get $${\left(\frac{b_n-1}{b_n}\right)}^{2018}+{\left(\frac{b_n-1}{b_n}\right)}^{2017}+\cdots +{\left(\frac{b_n-1}{b_n}\right)}^2+\frac{b_n-1}{b_n}-1345=0.$$Hence, $$(b_n-1{)}^{2018}+b_n(b_n-1{)}^{2017}+\cdots +b_n^{2016}(b_n-1{)}^2+b_n^{2017}(b_n-1)-1345b_n^{2018}=0.$$Thus, the $b_i$ are the roots of the polynomial $$(x-1{)}^{2018}+x(x-1{)}^{2017}+\cdots +x^{2016}(x-1{)}^2+x^{2017}(x-1)-1345x^{2018}=0.$$The coefficient of $x^{2018}$ is $2019-1346=673.$ The coefficient of $x^{2017}$ is $-1-2-\cdots -2018=-\frac{2018\cdot 2019}{2}.$ Therefore, the sum of the $b_i$ is $$\frac{2018\cdot 2019}{2\cdot 673}=\boxed{3027}.$$