One has $$ a+b<(1+o(1))n, $$ which is best possible in view of $1!n!|n!$. For the proof, assume that $a!b!|n!$, and consider the exponent of $2$ in the prime decomposition of the quotient $n!/(a!b!)$. This exponent is $$ \sum_{1\le k\le\log_2(a+b)} \big(\lfloor n/2^k\rfloor-\lfloor a/2^k\rfloor-\lfloor b/2^k\rfloor \big) = \sigma_1-\sigma_2, $$ where $$ \sigma_1 = \sum_{1\le k\le\log_2(a+b)} \big(\lfloor(a+b)/2^k\rfloor-\lfloor a/2^k\rfloor-\lfloor b/2^k\rfloor \big) $$ and $$ \sigma_2 = \sum_{1\le k\le\log_2(a+b)} \big(\lfloor(a+b)/2^k\rfloor - \lfloor n/2^k\rfloor \big). $$
We thus have $\sigma_1\ge\sigma_2$. On the other hand, it is well known that $\sigma_1\le\log_2(a+b)$ (for, every summand in $\sigma_1$ is either $0$ or $1$). The sum $\sigma_2$ can be estimated by $$ \sigma_2 \ge \sum_{1\le k\le\log_2(a+b)} \big((a+b)/2^k - n/2^k -1 \big) \ge a+b-n - \log_2(a+b) - O(1). $$ As a result, we get $$ a+b-n \le O(\log_2(a+b)), $$ implying the assertion.