3. (10 points) The first three numbers in a sequence are $1, 7, 8$, and each subsequent number is the remainder when the sum of the three preceding numbers is divided by 4. What is the sum of the first 2011 numbers in this sequence? $\qquad$ .
【Analysis】According to the problem, list the sequence as: $1, 7, 8, 0, 3, 3, 2, 0, 1, 3, 0, 0, 3, 3, 2, 0, 1, 3, 0, 0 \cdots$ It is evident that starting from the 4th number, every 8 numbers form a cycle. Since there are 3 numbers before the cycle, we need to subtract 3 from 2011 and then divide the result by 8 to find out how many groups there are, and then add them up to get the answer.

【Solution】Solution: The sequence: $1, 7, 8, 0, 3, 3, 2, 0, 1, 3, 0, 0, 3, 3, 2, 0, 1, 3, 0, 0 \cdots$
$(2011-3) \div 8=251$
$$
\begin{aligned}
& (0+3+3+2+0+1+3+0) \times 251+1+7+8 \\
= & 12 \times 251+16 \\
= & 3028
\end{aligned}
$$

Therefore, the answer is: 3028.