jmc

Let $q$ and $r$ be the remainder when $x$ is divided by 19, so $x=19q+r,$ where $0\le r\le 18.$ Then $$\begin{array}{rl}\lfloor x\rfloor -19\lfloor \frac{x}{19}\rfloor & =19q+r-19\lfloor \frac{19q+r}{19}\rfloor \\ & =19q+r-19\lfloor q+\frac{r}{19}\rfloor \\ & =19q+r-19q\\ & =r.\end{array}$$Thus, when $x$ is divided by 19, the remainder is 9. In other words, $x$ is 9 more than a multiple of 19.

Similarly, when $x$ is 9 more than a multiple of 89. Since 19 and 89 are relatively prime, $x$ is 9 greater than a multiple of $19\cdot 89=1691.$ Since $x$ is greater than 9, the smallest possible value of $x$ is $1691+9=\boxed{1700}.$