4. $n$ is a positive integer, and it is defined that: $n!=1 \times 2 \times \cdots \times$ $n$. Then the remainder when $1!\times 1+2!\times 2+\cdots+59!\times 59$ is divided by 2006 is ( ).
(A) 1
(B) 5
(C) 401
(D) 2005
4.D.

Notice that $2006=2 \times 17 \times 59$, so 2006160 !.
$$
\begin{array}{l}
\text { Also } 1! \times 1 + 2! \times 2 + \cdots + 59! \times 59 \\
=1! \times (2-1) + 2! \times (3-1) + \cdots + 59! \times (60-1) \\
=(2!-1!) + (3!-2!) + \cdots + (60!-59!) \\
=60!-1 .
\end{array}
$$

Therefore, the required remainder is 2005.