Example 2 The sequence $\left\{a_{n}\right\}$:
$1,1,2,1,1,2,3,1,1,2,1,1,2,3,4, \cdots$.
Its construction method is:
First, give $a_{1}=1$, then copy this item 1 and add its successor number 2, to get $a_{2}=1, a_{3}=2$;

Next, copy all the previous items $1,1,2$, and add the successor number 3 of 2, to get
$$
a_{4}=1, a_{5}=1, a_{6}=2, a_{7}=3 ;
$$

Next, copy all the previous items $1,1,2,1,1,2,3$, and add the successor number 4 of 3, to get the first 15 items as
$$
1,1,2,1,1,2,3,1,1,2,1,1,2,3,4 \text {; }
$$

Continue in this manner.
Try to find $a_{2000}$ and the sum of the first 2000 terms of the sequence $S_{2000}$.
From the construction method of the sequence $\left\{a_{n}\right\}$, it is easy to know that
$$
a_{1}=1, a_{3}=2, a_{7}=3, a_{15}=4, \cdots \cdots
$$

In general, we have $a_{2^{n}-1}=n$, meaning the number $n$ first appears at the $2^{n}-1$ term, and if $m=2^{n}-1+k\left(1 \leqslant k \leqslant 2^{n}-1\right)$, then
$$
\begin{array}{l}
a_{m}=a_{k} . \\
\text { Since } 2000=2^{10}-1+977, \\
977=2^{9}-1+466, \\
466=2^{8}-1+211, \\
211=2^{7}-1+84, \\
84=2^{6}-1+21, \\
21=2^{4}-1+6, \\
6=2^{2}-1+3,
\end{array}
$$

Thus, $a_{2000}=a_{977}=a_{466}=a_{211}=a_{84}$
$$
=a_{21}=a_{6}=a_{3}=2 \text {. }
$$

To find $S_{2000}$, we first calculate $S_{2^{n-1}}$.
By the construction method of the sequence $\left\{a_{n}\right\}$, we know that in the first $2^{n}-1$ terms, there are
1 $n$, 2 $(n-1)$, $2^{2}$ $(n-2)$, ..., 
$2^{k}$ $(n-k)$, ..., $2^{n-1}$ 1.
Thus, $S_{2^{n}-1}=n+2(n-1)+2^{2}(n-2)+\cdots+$
$$
2^{n-2} \times 2+2^{n-1} \times 1 \text {. }
$$

Therefore, $2 S_{2^{n}-1}=2 n+2^{2}(n-1)+2^{3}(n-2)+\cdots+$
$$
2^{n-1} \times 2+2^{n} \text {. }
$$

Subtracting the two equations, we get
$$
\begin{array}{l}
S_{2^{n-1}}=-n+\left(2+2^{2}+\cdots+2^{n-1}+2^{n}\right) \\
=2^{n+1}-(n+2) .
\end{array}
$$

When $m=2^{n}-1+k\left(1 \leqslant k \leqslant 2^{n}-1\right)$, we have
$$
\begin{array}{l}
S_{m}=S_{2^{n}-1}+\left[a_{\left(2^{n}-1\right)+1}+a_{\left(2^{n}-1\right)+2}+\cdots+a_{\left(2^{n}-1\right)+k}\right] \\
=S_{2^{n}-1}+\left(a_{1}+a_{2}+\cdots+a_{k}\right)=S_{2^{n}-1}+S_{k} .
\end{array}
$$

Therefore, $S_{2000}=S_{2^{10}-1}+S_{977}, S_{977}=S_{2^{9}-1}+S_{466}$,
$$
\begin{array}{l}
S_{466}=S_{2^{8}-1}+S_{211}, S_{211}=S_{2^{7}-1}+S_{84}, \\
S_{84}=S_{2^{6}-1}+S_{21}, S_{21}=S_{2^{4}-1}+S_{6}, S_{6}=8 .
\end{array}
$$

From equation (1), we get
$$
\begin{array}{l}
S_{2000}=\left(2^{11}-12\right)+\left(2^{10}-11\right)+\left(2^{9}-10\right)+ \\
\\
\left(2^{8}-9\right)+\left(2^{7}-8\right)+\left(2^{5}-6\right)+8 \\
=3952 .
\end{array}
$$