(1) We refer to the $k$-th 1 and the $2^{k-1}$ 2s that follow it, these $2^{k-1}+1$ terms as the $k$-th segment of the sequence. If the 1998th term is in the $k$-th segment, then $k$ is the smallest positive integer satisfying $k+\left(1+2+2^{2}+\cdots+2^{k-1}\right)=2^{k}+k-1 \geqslant 1998$.
$$
\begin{array}{l}
\because 2^{10}+10-1=10331998, \\
\therefore k=11 .
\end{array}
$$

Thus, $S_{1998}=2 \times 1998-11=3985$.
(2) If there exists $n \in \mathbf{N}$ such that $S_{n}=2001$, and the $n$-th term is in the $k$-th segment, then $2 n-k=2001$, which implies that $k$ is an odd number.
From (1), we know that $k \leqslant 11$, so $10001006>n$.
$$
\therefore \text{The } n \text{-th term should be in the 10th segment, i.e., } k=10, \text{ which contradicts that } k \text{ is an odd number. Therefore, } n \text{ does not exist.}
$$
Seven, given a sequence whose terms are 1 or 2, the first term is 1, and there are $2^{k-1}$ 2s 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.