2006
Problem - 1706
How many four-digit positive integers have at least one digit that is a $2$ or a $3$?
This can be solved using counting-by-opposite technique.
Therefore totally $9\times 10\times 10\times 10= 9000$ $4$-digit positive integers. Among which there are totally $7\times 8\times 8\times 8=3584$ $4$-digit numbers without $2$ and $3$. Therefore, the answer is $9000-3584=\boxed{5416}$.