4. (20 points) Given that the sum of two positive integers is 432, and the sum of the least common multiple (LCM) and the greatest common divisor (GCD) of these two positive integers is 7776. What is the product of these two positive integers?



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【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.