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One has $$ a+b<(1+o(1))n, $$ which is best possible in view of $1!n!|n!$. For the proof, assuming that $a!b!|n!$ with $a+b>n$, 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.

One has $$ a+b < n+O(\log n), $$ which is best possible in view of $1!n!|n!$. For the proof, assuming that $a!b!|n!$ with $a+b>n$, 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.

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.

One has $$ a+b<(1+o(1))n, $$ which is best possible in view of $1!n!|n!$. For the proof, assuming that $a!b!|n!$ with $a+b>n$, 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.

Here is a simple way to get the bound $$ a+b < (1.5+o(1))n. $$ Fix $\varepsilon>0$. If $a+b>(1.5+\varepsilon)n$, then either $\max\{a,b\}>n$, or $\min\{a,b\}>(0.5+\varepsilon)n$. In the former case $a!b!>n!$, hence $a!b!\mid n!$ is impossible. In the latter case, assuming that $n$ is sufficiently large, the interval $(0.5n,(0.5+\varepsilon)n)$ contains a prime. This prime divides both $a!$ and $b!$, whereas it enters $n!$ with the exponent just $1$ -- showing that $a!b!\mid n!$ cannot hold in this case, too.

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.