Given that \[ \frac 1{2!17!}+\frac 1{3!16!}+\frac 1{4!15!}+\frac 1{5!14!}+\frac 1{6!13!}+\frac 1{7!12!}+\frac 1{8!11!}+\frac 1{9!10!}=\frac N{1!18!} \] find the greatest integer that is less than $\frac N{100}.$
1. We start with the given equation:
   \[
   \frac{1}{2!17!} + \frac{1}{3!16!} + \frac{1}{4!15!} + \frac{1}{5!14!} + \frac{1}{6!13!} + \frac{1}{7!12!} + \frac{1}{8!11!} + \frac{1}{9!10!} = \frac{N}{1!18!}
   \]

2. Multiply both sides by \(19!\):
   \[
   19! \left( \frac{1}{2!17!} + \frac{1}{3!16!} + \frac{1}{4!15!} + \frac{1}{5!14!} + \frac{1}{6!13!} + \frac{1}{7!12!} + \frac{1}{8!11!} + \frac{1}{9!10!} \right) = 19! \cdot \frac{N}{1!18!}
   \]

3. Simplify each term on the left-hand side:
   \[
   \sum_{k=2}^9 \frac{19!}{k!(19-k)!} = \sum_{k=2}^9 \binom{19}{k}
   \]

4. The right-hand side simplifies to:
   \[
   19! \cdot \frac{N}{18!} = 19N
   \]

5. Therefore, we have:
   \[
   \sum_{k=2}^9 \binom{19}{k} = 19N
   \]

6. To find \(\sum_{k=2}^9 \binom{19}{k}\), we use the binomial theorem:
   \[
   \sum_{k=0}^{19} \binom{19}{k} = 2^{19}
   \]

7. Subtract the terms for \(k=0\) and \(k=1\) from the total sum:
   \[
   \sum_{k=2}^{19} \binom{19}{k} = 2^{19} - \binom{19}{0} - \binom{19}{1} = 2^{19} - 1 - 19 = 2^{19} - 20
   \]

8. Since \(\sum_{k=10}^{19} \binom{19}{k} = \sum_{k=0}^9 \binom{19}{19-k}\), we have:
   \[
   \sum_{k=2}^9 \binom{19}{k} = \frac{1}{2} \left( 2^{19} - 20 \right) = 2^{18} - 10
   \]

9. Therefore:
   \[
   19N = 2^{18} - 10
   \]

10. Solving for \(N\):
    \[
    N = \frac{2^{18} - 10}{19}
    \]

11. Calculate \(2^{18}\):
    \[
    2^{18} = 262144
    \]

12. Substitute back:
    \[
    N = \frac{262144 - 10}{19} = \frac{262134}{19} = 13796
    \]

13. Finally, find the greatest integer less than \(\frac{N}{100}\):
    \[
    \left\lfloor \frac{13796}{100} \right\rfloor = \left\lfloor 137.96 \right\rfloor = 137
    \]

The final answer is \(\boxed{137}\).