Question 1 Let $a_{1}, a_{2}, \cdots, a_{2018}$ be the roots of the polynomial
$$
x^{2018}+x^{2017}+\cdots+x^{2}+x-1345=0
$$

Calculate $\sum_{n=1}^{2018} \frac{1}{1-a_{n}}$.
(2018, Stanford University Mathematics Competition)
$$
\begin{array}{l}
\text { Let } P(x)=x^{2018}+x^{2017}+\cdots+x^{2}+x-1345 \text {. } \\
\text { Then } P(1)=1+1+\cdots+1+1-1345 \\
=2018-1345=673 \text {. } \\
\text { And } P^{\prime}(x) \\
=2018 x^{2017}+2017 x^{2016}+\cdots+2 x+1 \text {, }
\end{array}
$$

And $P^{\prime}(x)$
So $P^{\prime}(1)=2018+2017+\cdots+2+1$
$$
=\frac{2018(1+2018)}{2}=2019 \times 1009 \text {. }
$$
$$
\text { Then } \sum_{n=1}^{2018} \frac{1}{1-a_{n}}=\frac{P^{\prime}(1)}{P(1)}
$$
$$
\begin{array}{l}
=\frac{2019 \times 1009}{673} \\
=3 \times 1009=3027 .
\end{array}
$$