jmc

Prime factorize $8!$. $$\begin{array}{rl}8!& =8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\\ & =2^3\cdot 7\cdot (3\cdot 2)\cdot 5\cdot 2^2\cdot 3\cdot 2\\ & =2^7\cdot 3^2\cdot 5\cdot 7.\end{array}$$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, $N=2^3\cdot 3=\boxed{24}$.