3. $n$ is a positive integer, and it is defined that $n!=1 \times 2 \times \cdots \times n$ (for example, $1!=1,2!=1 \times 2$), then
$$
1!\times 1+2!\times 2+3!\times 3+\cdots+250!\times 250
$$
when divided by 2008, the remainder is ( ).
(A) 1
(B) 2
(C) 2007
(D) 2008
3.C.

Notice
$$
\begin{array}{l}
n! \times n = n! \times [(n+1)-1] \\
= n! \times (n+1) - n! = (n+1)! - n!.
\end{array}
$$

Then the original expression $=(2!-1!)+(3!-2!)+$
$$
\begin{aligned}
& (4!-3!)+\cdots+(251!-250!) \\
= & 251!-1=(251!-2008)+2007 .
\end{aligned}
$$

Since $2008 = 251 \times 8$, we have,
$$
2008 \text{ divides } (251! - 2008).
$$

Therefore, the remainder is 2007.