15. Divide 1983 into the sum of 12 positive integers to minimize the product of their factorials
15. The number of ways to divide 1983 into the sum of 12 positive integers is finite, so the set of products of their corresponding factorials is a non-empty finite subset of the set of natural numbers, and thus a minimum value exists.

Let $a_{1}, a_{2}, \cdots, a_{12}$ be positive integers. Suppose $a_{1}+a_{2}+\cdots+a_{12}=1983$, and $p=a!a_{2}!a_{3}! \cdots a_{12}!$ achieves the minimum value.
We will prove that for all $i, j=1,2, \cdots, 12$, we have $a_{i}-a_{j} \leqslant 1$.
If not, then there exist $a_{l}-a_{k}>1$ (where $l, k$ are two numbers among $1,2, \cdots, 12$), at this time $\left.a_{l}!a_{k}!=\left(a_{l}-1\right)!a_{l} \cdot a_{k}!>\left(a_{3}-1\right)!\left(a_{k}+1\right) \cdot a_{k}!=\left(a_{l}\right)-1\right)!\left(a_{k}+1\right)!\because\left(a_{l}-1\right)+\left(a_{k}+1\right)=a_{l}+a_{k}$, and $a_{l}!a_{k}!>\left(a_{l}-1\right)!\left(a_{k}+1\right)!$, so when $a_{l}-1$ replaces $a_{l}$ and $a_{k}+1$ replaces $a_{k}$ (with all others unchanged), the product will become smaller, which contradicts the fact that $p=a!a_{2}!\cdots a_{12}!$ achieves the minimum value. Therefore, the difference between any two numbers in $a_{1}, a_{2}, \cdots, a_{12}$ is at most 1.

Since $1983=12 \times 165+3$, when $a_{1}=a_{2}=\cdots=a_{9}=165, a_{10}=a_{11}=a_{12}=166$, $a_{1}!a_{2}!\cdots a_{12}!$ is minimized.