根据题意，有\[\begin{split}2017S&={\rm C}_{2017}^0-\left(1+\dfrac 1{2016}\right){\rm C}_{2016}^1+\left(1+\dfrac{2}{2015}\right){\rm C}_{2015}^2-\cdots+\left(1+\dfrac{1008}{1009}\right){\rm C}_{1009}^{1008}\\
&={\rm C}_{2017}^0-\left({\rm C}_{2016}^1+{\rm C}_{2015}^0\right)+\left({\rm C}_{2015}^2+{\rm C}_{2014}^1\right)-\cdots+\left({\rm C}_{1009}^{1008}+{\rm C}_{1008}^{1007}\right)\\
&=S_{2017}-S_{2015}
,\end{split}\]其中\[\begin{split}S_{2017}&={\rm C}_{2017}^0-{\rm C}_{2016}^1+{\rm C}_{2015}^2-\cdots+{\rm C}_{1009}^{1008},\\ S_{2015}&={\rm C}_{2015}^0-{\rm C}_{2014}^1+{\rm C}_{2013}^2-\cdots-{\rm C}_{1008}^{1007},\end{split}\]类似的定义\[\begin{split}S_{2k}&={\rm C}_{2k}^0-{\rm C}_{2k-1}^1+{\rm C}_{2k-2}^2-\cdots+(-1)^k{\rm C}_k^k,\\
S_{2k+1}&={\rm C}_{2k+1}^0-{\rm C}_{2k}^1+{\rm C}_{2k-1}^2-\cdots+(-1)^k{\rm C}_{k+1}^k,\end{split}\]那么由组合恒等式\[{\rm C}_n^m={\rm C}_{n-1}^{m-1}+{\rm C}_{n-1}^m,\]可得\[S_{2k+1}-S_{2k}=-S_{2k-1},S_{2k+2}-S_{2k+1}=-S_{2k},\]因此有\[S_{n+2}=S_{n+1}-S_n,\]于是数列 $\{S_n\}$ 是周期为 $6$ 的数列．考虑到 $S_1=1$，$S_2=0$，因此\[S_n:\underbrace{1,0,-1,-1,0,1,1},\underbrace{1,0,-1,-1,0,1,1},\cdots,\]因此\[S_{2017}=S_1=1,S_{2015}=S_5=0,\]进而所求代数式的值为 $\dfrac{1}{2017}$．