Example 6 Arrange the terms of the arithmetic sequence $2,6,10,14, \cdots, 2006$ tightly together to form a “large number”: $A=261014 \cdots 2006$. Find the remainder when $A$ is divided by 9.
First, we prove a lemma.
Lemma When a positive integer $n$ is arbitrarily divided into several groups, each group being a number formed by several consecutive digits of $n$, let the numbers corresponding to each group (from right to left) be $A_{1}, A_{2}, \cdots, A_{t}$, then
$$
n \equiv A_{1}+A_{2}+\cdots+A_{t}(\bmod 9) .
$$

Proof Let $A_{1}, A_{2}, \cdots, A_{t}$ be $r_{1}, r_{2}, \cdots, r_{t}$-digit numbers, respectively.
$$
\begin{array}{l}
\text { Let } s_{i}=r_{1}+r_{2}+\cdots+r_{i} \text {. Then } \\
n=\overline{A_{t} A_{t-1} \cdots A_{1}} \\
=A_{t} \cdot 10^{s_{t}}+A_{t-1} \cdot 10^{s_{t-1}}+\cdots+A_{1} \cdot 10^{s_{1}} \\
\equiv A_{t} \cdot 1^{s_{t}}+A_{t-1} \cdot 1^{s_{t-1}}+\cdots+A_{1} \cdot 1^{s_{1}} \\
\equiv A_{t}+A_{t-1}+\cdots+A_{1}(\bmod 9) .
\end{array}
$$

The lemma is proved.
By the lemma, we know
$$
\begin{array}{l}
A \equiv 2+6+10+\cdots+2006=1004 \times 502 \\
\equiv(1+4) \times(5+2)=35 \equiv 8(\bmod 9) .
\end{array}
$$

Therefore, the remainder of $A$ divided by 9 is 8.