jmc

By Descartes' Rule of Signs, none of the polynomials has a positive root, and each one has exactly one negative root. Furthermore, each polynomial is positive at $x=0$ and negative at $x=-1,$ so each real root lies between $-1$ and 0. Also, each polynomial is increasing on the interval $(-1,0).$

Let $r_A$ and $r_B$ be the roots of the polynomials in options A and B, respectively, so $$r_A^{19}+2018r_A^{11}+1=r_B^{17}+2018r_B^{11}+1=0,$$so $r_A^{19}=r_B^{17}.$ Since $r_A\in (-1,0),$ $r_B^{17}=r_A^{19}>r_A^{17},$ so $r_B>r_A.$

Similarly, let $r_C$ and $r_D$ be the roots of the polynomials in options C and D, respectively, so $$r_C^{19}+2018r_C^{13}+1=r_D^{17}+2018r_D^{13}+1=0,$$so $r_C^{19}=r_D^{17}.$ Since $r_C\in (-1,0),$ $r_D^{17}=r_C^{19}>r_C^{17},$ so $r_D>r_C.$

Since $$r_B^{17}+2018r_B^{11}+1=r_D^{17}+2018r_D^{13}+1=0,$$we have that $r_B^{11}=r_D^{13}.$ Since $r_D\in (-1,0),$ $r_B^{11}=r_D^{13}>r_D^{11},$ so $r_B>r_D.$

Therefore, the largest root must be either $r_B$ or the root of $2019x+2018=0,$ which is $-\frac{2018}{2019}.$

Let $f(x)=x^{17}+2018x^{11}+1,$ so $f(r_B)=0.$ Note that $$f\left(-\frac{2}{3}\right)=-\frac{2^{17}}{3^{17}}-2018\cdot \frac{2^{11}}{3^{11}}+1.$$We claim that $2018\cdot 2^{11}>3^{11}.$ Since $2^2>3,$ $2^{22}>3^{11}.$ Then $$2018\cdot 2^{11}=1009\cdot 2^{22}>3^{11}.$$From $2018\cdot 2^{11}>3^{11},$ $2018\cdot \frac{2^{11}}{3^{11}}>1,$ so $$f\left(-\frac{2}{3}\right)=-\frac{2^{17}}{3^{17}}-2018\cdot \frac{2^{11}}{3^{11}}+1<0.$$Since $f(x)$ is an increasing function, we can conclude that $r_B>-\frac{2}{3}>-\frac{2018}{2019}.$ Therefore, the answer is $\boxed{\text{(B)}}.$