258. Find all natural $n$ for which the number $10^{n}-1$ is divisible by 101.
$\triangle$ Let's transform this number:

$$
10^{n}-1=999 \ldots 9 \text { ( } n \text { nines). }
$$

For which number of nines is the resulting number divisible by 101? Checking shows that the numbers 9, 99, and 999 are not divisible by 101, but the number 9999 is divisible by 101, since 9999 is divisible by 1111, and 1111 is divisible by 101.

It appears that the given number is divisible by 101 only when $n$ is divisible by 4. Let's consider two cases.

1) Suppose $n$ is divisible by 4: $n=4 k(k \in N)$. We get:

$$
10^{n}-1=10^{4 k}-1=\left(10^{2}\right)^{2 k}-1
$$

and the last number is divisible by $10^{2}+1=101$ (see § 7, third divisibility).

2) Suppose $n$ is not divisible by 4. Divide $n$ by 4 with a remainder:

$$
n=4 k+r
$$

where $k$ is a non-negative integer, and $r$ is an integer satisfying the inequality $1 \leqslant r \leqslant 3$. Transform the given number:

$$
\begin{gathered}
10^{n}-1=10^{4 k+r}-1=10^{r} \cdot 10^{4 k}-1=10^{r}\left(\left(10^{4 k}-1\right)+1\right)-1= \\
=10^{r}\left(10^{4 k}-1\right)+\left(10^{r}-1\right) .
\end{gathered}
$$

The first term of the resulting sum is divisible by 101, but the second term is not, since for $r=1,2,3$ the number $10^{r}-1$ takes the values 9, 99, and 999, respectively. Therefore, the entire sum is not divisible by 101.

Answer: all $n$ that are divisible by 4.