One, (20 points) Given
$$
f(x)=(x-1)(x-2) \cdots \cdots(x-2004) \text {. }
$$

Ask: How many real numbers $x$ satisfy $|f(x)|=1$?
一、As shown in Figure 7, draw a sketch of the function $f(x)$.
From the graph, we know that between 1 and 2004, for every 2 consecutive integers, there are 2 real numbers such that $|f(x)|=1$.

In the intervals $(0,1)$ and $(2004,+\infty)$, there is 1 real number $x$ in each interval such that $|f(x)|=1$.

Therefore, there are a total of $2003 \times 2+1+1=4008$ real numbers $x$ such that $|f(x)|=1$.