3. (25 points) Given that $a$ and $b$ are coprime positive integers, satisfying $a+b=2005$. Let $[x]$ denote the integer part of the number $x$, and let
$$
\begin{array}{l}
A=\left[\frac{2005 \times 1}{a}\right]+\left[\frac{2005 \times 2}{a}\right]+\cdots+\left[\frac{2005 \times a}{a}\right], \\
B=\left[\frac{2005 \times 1}{b}\right]+\left[\frac{2005 \times 2}{b}\right]+\cdots+\left[\frac{2005 \times b}{b}\right] .
\end{array}
$$

Try to find the value of $A+B$.
3. It is evident that the last terms of the two sums $A$ and $B$ are both 2005. Consider the remaining terms.
Since $(a, b)=1, a+b=2005$, it follows that
$(a, 2005)=1, (b, 2005)=1$.
Therefore, for $1 \leqslant m \leqslant a-1, 1 \leqslant n \leqslant b-1, \frac{2005 m}{a}$ and $\frac{2005 n}{b}$ are not integers, and they are not equal.

In fact, if $\frac{2005 m}{a}=\frac{2005 n}{b}$, then $\frac{a}{b}=\frac{m}{n}$, and $1 \leqslant m \leqslant a-1, 1 \leqslant n \leqslant b-1$, which contradicts the fact that $a$ and $b$ are coprime.
Now, consider a set of intervals of length 1 on the number line:
$(1,2),(2,3), \cdots,(2003,2004)$, each of which is called a "unit interval".
Since $\frac{2005}{a}>1$, then $\frac{2005(a-1)}{a}k a, 2005 n>k b$

and $2005 mk$.
By adding the two equations in (2), we get $m+n<k+1$.
Therefore, $k<m+n<k+1$, which contradicts the fact that the interval $(k, k+1)$ does not contain any integer. Thus,
$$
\begin{array}{r}
\frac{2005 \times 1}{a}, \frac{2005 \times 2}{a}, \cdots, \frac{2005 \times(a-1)}{a} \\
\text { and } \quad \frac{2005 \times 1}{b}, \frac{2005 \times 2}{b}, \cdots, \frac{2005 \times(b-1)}{b}
\end{array}
$$

contain a total of $a+b-2(=2003)$ numbers, each falling into different intervals $(1,2)$, $(2,3), \cdots,(2003,2004)$. Each interval contains exactly one of these numbers. When a number $x$ belongs to the interval $(k, k+1)$, we have $[x]=k$. Therefore,
$$
\begin{array}{l}
A+B=(1+2+3+\cdots+2003)+2005+2005 \\
=\frac{2003 \times 2004}{2}+4010=2011016 .
\end{array}
$$
(Provided by Ping Sheng)