jmc

Multiplying both sides by $19!$ yields: $$\frac{19!}{2!17!}+\frac{19!}{3!16!}+\frac{19!}{4!15!}+\frac{19!}{5!14!}+\frac{19!}{6!13!}+\frac{19!}{7!12!}+\frac{19!}{8!11!}+\frac{19!}{9!10!}=\frac{19!N}{1!18!}.$$ $$\left(\genfrac{}{}{0ex}{}{19}{2}\right)+\left(\genfrac{}{}{0ex}{}{19}{3}\right)+\left(\genfrac{}{}{0ex}{}{19}{4}\right)+\left(\genfrac{}{}{0ex}{}{19}{5}\right)+\left(\genfrac{}{}{0ex}{}{19}{6}\right)+\left(\genfrac{}{}{0ex}{}{19}{7}\right)+\left(\genfrac{}{}{0ex}{}{19}{8}\right)+\left(\genfrac{}{}{0ex}{}{19}{9}\right)=19N.$$ Recall the Combinatorial Identity $2^{19}={\sum }_{n=0}^{19}\left(\genfrac{}{}{0ex}{}{19}{n}\right)$. Since $\left(\genfrac{}{}{0ex}{}{19}{n}\right)=\left(\genfrac{}{}{0ex}{}{19}{19-n}\right)$, it follows that ${\sum }_{n=0}^9\left(\genfrac{}{}{0ex}{}{19}{n}\right)=\frac{2^{19}}{2}=2^{18}$. Thus, $19N=2^{18}-\left(\genfrac{}{}{0ex}{}{19}{1}\right)-\left(\genfrac{}{}{0ex}{}{19}{0}\right)=2^{18}-19-1=(2^9{)}^2-20=(512{)}^2-20=262124$. So, $N=\frac{262124}{19}=13796$ and $\lfloor \frac{N}{100}\rfloor =\boxed{137}$.