Seven, (Full marks 12 points) Given a sequence where each term is 1 or 2, the first term is 1, and there are $2^{k-1}$ twos between the $k$-th 1 and the $(k+1)$-th 1, i.e., $1,2,1,2,2,1,2,2,2,2,1,2,2,2,2,2,2,2,2,1, \cdots$
(1) Find the sum of the first 1998 terms of the sequence, $S_{1998}$;
(2) Does there exist a positive integer $n$ such that the sum of the first $n$ terms of the sequence, $S_{n}=2001$? If $n$ exists, find the value of $n$; if $n$ does not exist, prove your conclusion.
For the $k$-th segment, we take the $k$-th 1 and then follow it with $2^{k-1}$ 2's, making the $2^{k-1}+i$-th term the $k$-th segment of the sequence.
Suppose the 1958th term is in the $k$-th segment. Then $k$ is the smallest positive integer satisfying
$$
\dot{\kappa}+\left(1+2+2^{2}+\cdots+2^{k-1}\right)=2^{k}+k-1 \geqslant 1998
$$
\[
\begin{array}{l}
\because \quad 2^{10}+10-1=10331998, \\
\therefore k=11 .
\end{array}
\]

Thus, $S_{1998}=2 \times 1998-11=3985$.
(2) If there exists $n \in \mathbb{N}$ such that $S_{n}=2001$. Suppose the $n$-th term is in the $k$-th segment. Then we have
$$
2 n-k=2001 \text {. }
$$

This implies that $k$ is an odd number.
From (1), we know $k \leqslant 11$, so $10001006>n,
\end{aligned}
$$
$\therefore$ the $n$-th term should be in the 10th segment, i.e., $k=10$, which contradicts the fact that $k$ is an odd number. Therefore, $n$ does not exist.
(Provided by Cao Youwen)