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)}} $.