Solution:

The answer is (A). Observe that $N=10^{2024}+1$ and that a divisor of it of the required type is written as $10^{n}+1$ for some $1<n<2024$. Let us therefore assume that $10^{n}+1$ divides $N$ and write
$$N=10^{2024}+1=\left(10^{n}+1\right)\left(10^{2024-n}\right)+r_{1},$$
from which $r_{1}=-\left(10^{2024-n}-1\right)$. Since $N$ is divisible by $10^{n}+1$, $r_{1}$ is also divisible by $10^{n}+1$. Repeating the same procedure, we obtain
$$-r_{1}=10^{2024-n}-1=\left(10^{n}+1\right)\left(10^{2024-2 n}\right)+r_{2},$$
from which $r_{2}=-\left(10^{2024-2 n}+1\right)$ is divisible by $10^{n}+1$ (essentially we are performing a sort of long division, in which however we allow the remainder to be negative).

Iterating this reasoning we obtain that $10^{2024-2 k n}+1$ is a multiple of $10^{n}+1$ for every $k$ such that $2024-2 k n$ is positive, and similarly $10^{2024-(2 k+1) n}-1$ must be a multiple of $10^{n}+1$ for every $k$ such that $2024-(2 k+1) n$ is positive. In particular, by choosing $k$ appropriately we can make $2024-2 k n$ or $2024-(2 k+1) n$ positive and less than or equal to $n$ (this essentially amounts to carrying out the division with remainder between 2024 and $n$): we then obtain that $10^{n}+1$ divides a number of the form $10^{a} \pm 1$ with $1 \leq a \leq n$. If $a$ were strictly less than $n$, clearly $10^{a} \pm 1$ would be less than $10^{n}+1$, and hence could not be a multiple of it. It must therefore happen that $a=n$ and that the sign is positive, which occurs only if $n=2024-2 k n$, that is $2024=(2 k+1) n$. Thus $n$ must be a proper divisor of 2024 with the property that 2024/n is odd, that is $n=2^{3} d$, where $d$ is a proper divisor of $2024 / 2^{3}=253=11 \cdot 23$. There are therefore 3 possibilities: $n=2^{3}, n=2^{3} \cdot 11, n=2^{3} \cdot 23$.