Question 3. The sum of several natural numbers is 1983. When is their product the largest, and what is the maximum value?

Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
Since $\frac{1983}{\left[\frac{1983}{e}\right]}$ and $\frac{1983}{\left[\frac{1983}{e}\right]+1}$ are not integers, the solution method from the previous problem cannot be applied. A completely different approach is needed.

Assume that the natural numbers $a_{1}, a_{2}, \cdots, a_{n}$ (where $\mu$ is also unknown) sum to 1983, and their product $P$ is maximized. Since for $x \geqslant 4$,
$$
2(x-2) \geqslant x,
$$

we can assume that $a_{i} \leq 3$ for $i=1,2, \cdots, n$, otherwise we can replace two numbers $a_{1}, a_{2}$ with one number $1 + a_{1}$, keeping the sum as 1983 while increasing the product $P$.
Thus, $a_{i}$ can only be 2 or 3 for $i=1,2, \cdots, n$. Note that
$$
2 \times 2 \times 2 < 3 \times 3,
$$

so there can be at most two 2s in $a_{1}, a_{2}, \cdots, a_{n}$, otherwise we can replace $2, 2, 2$ with 3, 3, keeping the sum as 1983 while increasing the product $P$. Therefore, the maximum product $P = 2^x \cdot 3^y$, where $y$ is a non-negative integer, and $x = 0, 1, 2$.
And
$$
2x + 3y = 1983
$$

If $x = 1$ or 2, (1) has no integer solutions. When $x = 0$, from (1) we get $y = 661$. Therefore, the maximum product is $3^{661}$, and it is achieved when all the addends are 3. The addends do not necessarily have to be all equal. For example, if 1983 is replaced with 1084, the answer is: the maximum product is $2^2 \times 3^{360}$, and it is achieved when 660 addends are 3 and 2 addends are 2 (or 1 addend is 4). This further demonstrates that applying the solution methods from problems 1 and 2 to problem 3 is incorrect.