jmc

We can write $$f(x)=(x-r_1)(x-r_2)\cdots (x-r_{2017})$$and $$P(z)=k\prod _{j=1}^{2007}\left(z-\left(r_j+\frac{1}{r_j}\right)\right)$$for some nonzero constant $k.$

We want to compute $$\frac{P(1)}{P(-1)}=\frac{{\prod }_{j=1}^{2007}\left(1-\left(r_j+\frac{1}{r_j}\right)\right)}{{\prod }_{j=1}^{2007}\left(-1-\left(r_j+\frac{1}{r_j}\right)\right)}=\frac{{\prod }_{j=1}^{2007}(r_j^2-r_j+1)}{{\prod }_{j=1}^{2007}(r_j^2+r_j+1)}.$$Let $\alpha$ and $\beta$ be the roots of $x^2+x+1=0,$ so $$x^2+x+1=(x-\alpha )(x-\beta ).$$Then $$x^2-x+1=(x+\alpha )(x+\beta ).$$Also, $(\alpha -1)({\alpha }^2+\alpha +1)={\alpha }^3-1=0,$ so ${\alpha }^3=1.$ Similarly, ${\beta }^3=1.$ Thus, $$\begin{array}{rl}\prod _{j=1}^{2007}(r_j^2-r_j+1)& =\prod _{j=1}^{2007}(r_j+\alpha )(r_j+\beta )\\ & =\prod _{j=1}^{2007}(-\alpha -r_j)(-\beta -r_j)\\ & =f(-\alpha )f(-\beta )\\ & =(-{\alpha }^{2007}+17{\alpha }^{2006}+1)(-{\beta }^{2007}+17{\beta }^{2006}+1)\\ & =(17{\alpha }^2)(17{\beta }^2)\\ & =289.\end{array}$$Similarly, $$\begin{array}{rl}\prod _{j=1}^{2007}(r_j^2+r_j+1)& =\prod _{j=1}^{2007}(r_j-\alpha )(r_j-\beta )\\ & =\prod _{j=1}^{2007}(\alpha -r_j)(\beta -r_j)\\ & =f(\alpha )f(\beta )\\ & =({\alpha }^{2007}+17{\alpha }^{2006}+1)({\beta }^{2007}+17{\beta }^{2006}+1)\\ & =(17{\alpha }^2+2)(17{\beta }^2+2)\\ & =289{\alpha }^2{\beta }^2+34{\alpha }^2+34{\beta }^2+4\\ & =259.\end{array}$$Therefore, $$\frac{P(1)}{P(-1)}=\boxed{\frac{289}{259}}.$$