2.98 Let $S$ be the set of all non-negative integers in decimal notation that have at most $n$ digits, and let $S_{k}$ consist of those elements of $S$ whose digits sum to less than $k$. For which $n$ does there exist a $k$ such that $|S|=2\left|S_{k}\right|$?
[Solution] For any $n$-digit number $A=\overline{a_{1} a_{2} \cdots a_{n}}\left(0 \leqslant a_{i} \leqslant 9, i=1,2, \cdots\right.$, $n)$, corresponding to
$$
A \rightarrow B=\overline{b_{1} b_{2} \cdots b_{n}}
$$

is a bijection from the set of all non-negative integers with at most $n$ digits to itself, where $b_{i}=9-a_{i}, i=1,2, \cdots, n$. If we denote $d(A)=a_{1}+a_{2}+\cdots+a_{n}$, then
$$
d(A)+d(B)=9 n \text {. }
$$

From this, for any $0 \leqslant k \leqslant 9 n$, we have
$$
\left|\left\{A \left\lvert\, d(A)\frac{9 n}{2}\right.\right\}\right| .
$$

When $n$ is odd, $\frac{9 n}{2}$ is not an integer, so the union of the sets on both sides of (1) is $S$. Therefore, when $k=\left[\frac{9 n}{2}\right]+1$, $|S|=2\left|S_{k}\right|$.

When $n$ is even, $\frac{9 n}{2}$ is an integer. When $k=\frac{9 n}{2}$, $|S|>2\left|S_{k}\right|,|S|<$ $2\left|S_{k+1}\right|$. In this case, there does not exist a $k$ that meets the requirement.
In summary, a natural number $k$ that meets the requirement exists if and only if $n$ is odd.