jmc

We can write $$\begin{array}{rl}\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+\cdots +\frac{1}{2016}& =\frac{2017-1}{1}+\frac{2017-2}{2}+\frac{2017-3}{3}+\cdots +\frac{2017-2016}{2016}\\ & =\frac{2017}{1}-1+\frac{2017}{2}-1+\frac{2017}{3}-1+\cdots +\frac{2017}{2016}-1\\ & =\frac{2017}{1}+\frac{2017}{2}+\frac{2017}{3}+\cdots +\frac{2017}{2016}-2016\\ & =2017\left(\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{2016}\right)+1\\ & =2017\left(\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{2016}+\frac{1}{2017}\right).\end{array}$$Therefore, $$\frac{\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+\cdots +\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots +\frac{1}{2017}}=\boxed{2017}.$$