Example 3 The sequence $\left\{a_{n}\right\}$:
$1,1,2,1,1,2,3,1,1,2,1,1,2,3,4, \cdots$, is called a "fractal sequence", and its construction method is as follows:

First, give $a_{1}=1$, then copy this item 1 and add its successor number 2, to get $a_{2}=1, a_{3}=2$;

Then 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 $1,1,2,1,1,2,3,1,1,2,1,1,2,3,4$; and so on.
Try to find $a_{2000}$ and the sum of the first 2000 terms of the sequence $S_{2000}$. (2000, Nanchang City High School Mathematics Competition)
Solving the construction method of the sequence $\left\{a_{n}\right\}$, we easily know
$$
a_{1}=1, a_{3}=2, a_{7}=3, a_{15}=4, \cdots \cdots
$$

Generally, $a_{2^{n-1}}=n$, which means 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 $a_{m}=a_{k}$.
Given $2000=2^{10}-1+977$,
$977=2^{9}-1+466,466=2^{8}-1+211$,
$211=2^{7}-1+84,84=2^{6}-1+21$,
$21=2^{4}-1+6$,
thus $a_{2000}=a_{977}=a_{466}=a_{211}$
$$
=a_{84}=a_{21}=a_{6}=2 \text {. }
$$

To find $S_{2000}$, we first calculate $S_{2^{n}-1}$.
From the construction method of the sequence $\left\{a_{n}\right\}$, we know that in the first $2^{n}-1$ terms of the sequence, there is exactly 1 $n$, 2 $n-1$s, $2^{2}$ $n-2$s, $\cdots, 2^{k}$ $n-k$s, $\cdots, 2^{n-1}$ 1s.
Thus, $S_{2^{n-1}}=n+2(n-1)+2^{2}(n-2)+$
$$
\begin{array}{c}
\cdots+2^{n-2} \times 2+2^{n-1} \times 1, \\
2 S_{2^{n-1}}=2 n+2^{2}(n-1)+2^{3}(n-2)+ \\
\cdots+2^{n-1} \times 2+2^{n} .
\end{array}
$$

Subtracting the above 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 $m=2^{n}-1+k\left(1 \leqslant k \leqslant 2^{n}-1\right)$,
$$
\begin{array}{l}
S_{m}=S_{2^{n}-1}+\sum_{i=1}^{k} a_{\left(2^{n}-1\right)+i} \\
=S_{2^{n}-1}+\left(a_{1}+a_{2}+\cdots+a_{k}\right)=S_{2^{n}-1}+S_{k} .
\end{array}
$$

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

Substituting into equation (1), we get
$$
\begin{aligned}
S_{2000}= & \left(2^{11}-12\right)+\left(2^{10}-11\right)+\left(2^{9}-10\right)+ \\
& \left(2^{8}-9\right)+\left(2^{7}-8\right)+\left(2^{5}-6\right)+6 \\
= & 3950
\end{aligned}
$$