Since we are asked for the number of positive $4$-digit integers with at least $2$ or $3$ in it, we can find this by finding the total number of $4$-digit integers and subtracting off those which do not have any $2$s or $3$s as digits.
The total number of $4$-digit integers is $9 \cdot 10 \cdot 10 \cdot 10 = 9000$, since we have $10$ choices for each digit except the first (which can't be $0$).
Similarly, the total number of $4$-digit integers without any $2$ or $3$ is $7 \cdot 8 \cdot 8 \cdot 8 ={3584}$.
Therefore, the total number of positive $4$-digit integers that have at least one $2$ or $3$ is $9000-3584=\boxed{\textbf{(E) }5416}.$