Answer: 137

Solution:

Multiply both sides by $19!.$ Each term on the left becomes $\frac{19!}{k!\,(19 - k)!} = \binom{19}{k}$ for $k = 2, \ldots, 9,$ while the right side becomes $19N.$

Since $\binom{19}{k} = \binom{19}{19 - k},$ the first half of the binomial row sums to $\sum_{k=0}^{9} \binom{19}{k} = \frac{2^{19}}{2} = 2^{18},$ so $$\begin{aligned} \sum_{k=2}^{9}\binom{19}{k} &= 2^{18} - \binom{19}{0} \\ &\quad {}- \binom{19}{1} \\ &= 262144 - 20 \\ &= 262124. \end{aligned}$$

Hence $N = \frac{262124}{19} = 13796,$ so $\frac{N}{100} = 137.96,$ and the greatest integer less than this is $137.$