Problem 855

【Analysis】This problem involves the sum of divisors and multiples, not the product. Therefore, they need to be represented separately. List the equation relationships. Let $A=a N, B=b N$, i.e., $a N+b N=432, \mathbf{N}+a b N=7776$. Both equations contain $N$, which can be canceled out to find the relationship between $a$ and $b$. Determine the specific values based on the requirement that all numbers are positive integers.

【Solution】Solution: Let the two positive integers be $A, B$. Their common divisor is $N. A=a N, B=b N$. $(a, b$ are coprime) The least common multiple is $a b N$.
$a N+b N=N(a+b)=432$, factorize $432=2^{4} \times 3^{3}$
$\mathbf{N}+a b N=N(a b+1)=7776$, factorize $7776=2^{5} \times 3^{5}$
$\frac{\mathrm{N}(\mathrm{a}+\mathrm{b})}{\mathrm{N}(\mathrm{ab}+1)}=\frac{432}{7776}$, solving gives $a b+1=18(a+b)$,
Rearranging gives:
$$b=\frac{18 a-1}{a-18}=\frac{18(a-18)+323}{a-18}=18+\frac{323}{a-18},$$
$\because a, b$ are integers, $\therefore \frac{323}{a-18}$ is an integer, so $a-18$ must be a divisor of 323.
$$323=17 \times 19,$$

When $a-18=17$, $b=18+19=37, a=35$,
When $a-18=19$, $b=18+18=35, b=37$,
$$\begin{array}{l} \therefore a b=35 \times 37=1295, \\ N(1295+1)=7776, \\ N=6, \\ A B=a b N^{2}=1295 \times 36=46620, \end{array}$$

In conclusion, the answer is: 46620.