The sum and product of the first 51.1995 integers are both equal to F $\cdot$ 1996. What is the sum of the absolute values of these 1995 integers? Please prove your conclusion.

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Solution: Let these 1995 integers be denoted as $x_{1}, x_{2}, \cdots, x_{1995}$. Then
$$
\left\{\begin{array}{l}
x_{1}+x_{2}+\cdots+x_{1995}=1996, \\
x_{1} x_{2} \cdots \cdots x_{1995}=1996=2^{2} \times 499 .
\end{array}\right.
$$
(1) From (1), we know that among these 1995 numbers, there are an even number of odd numbers and an odd number of even numbers.
(2) From (2), we know that among these 1995 numbers, there are an even number of negative numbers and an odd number of positive numbers.
(3) From (2), we know that among these 1995 numbers, there can only be one or two even numbers. Combining this with (1), we know that there can only be one even number. This even number can be $\pm 4$ or $\pm 1996$.

Next, we prove that it is impossible to take $\pm 1996$. Otherwise, if $x_{1995} = 1996$, the other numbers can only be +1 or -1, and let their counts be $k_{1}$ and $k_{2}$, respectively. From (2), we know that $k_{1}$ and $k_{2}$ are both non-negative even numbers. From (1) and (2), we get
$$
\left\{\begin{array}{l}
k_{1}-k_{2}+1996=1996, \\
k_{1}+k_{2}=1994 .
\end{array}\right.
$$

It is easy to see that the solution to this system of equations is $k_{1}=k_{2}=997$, which are not even numbers. Therefore, the assumption is not valid.

If $x_{1995} = -1996$, since the sum of the absolute values of the other numbers is 1994, equation (1) has no solution, which is still impossible.

Since 2 and 499 are both prime numbers, these 1995 numbers can only be $\pm 1, \pm 4, \pm 499$.

If $x_{1995} = -499$ and $x_{1994} = 4a$, where $a$ is +1 or -1, and the rest are $k_{1}$ +1s and $k_{2}$ -1s, then
$$
k_{1}-k_{2}+4a-499=1996.
$$

Since $k_{1}+k_{2}=1993$, it is clear that the left side is less than 1996, so the equation has no solution. Therefore, we can only take $x_{1995} = 499$. In this case, we have
$$
k_{1}-k_{2}+4a+499=1996.
$$

That is,
$$
\left\{\begin{array}{l}
k_{1}-k_{2}+4a=1497, \\
k_{1}+k_{2}=1993 .
\end{array}\right.
$$
Solving this, we get $k_{1}=1745-2a, k_{2}=248+2a$.
From (2), we know that the number of negative numbers is even, so we must have $a=1$ (otherwise, there will be $k_{2}+1$ negative numbers), i.e., $k_{1}=1743, k_{2}=250$. The sum of the absolute values is $1743+250+4+499=2496$.
Note: The editor made significant modifications to the solution.
(Fu Jie, Huayuan County National Middle School, Hunan Province, 416400)