The numbers between $999$ and $10000$ are all $4$-digit numbers.
The number of $4$ - digit numbers formed by digits
$0,2,3,6,7,8$ is ${ }^{6} P _{4}=360$.
But here those numbers are also involved which begin from $0 .$
So, we take those numbers as three-digit numbers.
Taking initial digit $0$ , the number of ways to fill remaining $3$ places from five digits
$2,3,6,7,8$ are ${ }^{5} P _{3}=60$
So, the required numbers $=360-60=300$.