Prime factorize $8!$. \begin{align*} 8!&= 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2 \\ &=2^3\cdot 7\cdot(3\cdot2)\cdot5\cdot 2^2\cdot 3\cdot 2\\ &=2^7\cdot 3^2\cdot 5 \cdot 7.\end{align*} Since $ N^2 $ is a divisor of $8!$, the exponents in the prime factorization of $ N^2 $ must be less than or equal to the corresponding exponents in the prime factorization of 8!. Also, because $ N^2 $ is a perfect square, the exponents in its prime factorization are all even. Therefore, the largest possible value of $ N^2 $ is $ 2^6\cdot 3^2 $. Square rooting both sides, $ N=2^3\cdot 3=\boxed{24} $.