Any number of the form $2^{2022}p$, where $p>2^{2022}+2^{4044}$ is a prime, works. Clearly there are infinitely many of them. To check these numbers are not southern, the divisors are arranged as follows: $1, 2, 4, \cdots, 2^{2022}, p, 2p, 4p, \cdots, 2^{2022}p$. Assume FTSOC that $p-2^{2022} | 2^{2022}p$. It also divides $2^{2022}p-2^{4044}$, so it divides their difference which is $2^{4044}$, but this is a contradiction since $p-2^{2022}>2^{4044}$. For similar reasons, $2^kp$ is not southern, where $k<2022$. It can be seen that $2^n$ is southern, since all differences of adjacent divisors are of the form $2^k$ with $k<n$. Thus, all divisors of the form $2^k$ with $k\in[1,2,\cdots,2022]$ are southern, and there are $2022$ of them. This solves part (b), which also implies part (a), solving the problem.

onlyvipers wrote: Any number of the form $2^{2022}p$, where $p>2^{2022}+2^{4044}$ is a prime, works. Clearly there are infinitely many of them. To check these numbers are not southern, the divisors are arranged as follows: $1, 2, 4, \cdots, 2^{2022}, p, 2p, 4p, \cdots, 2^{2022}p$. Assume FTSOC that $p-2^{2022} | 2^{2022}p$. It also divides $2^{2022}p-2^{4044}$, so it divides their difference which is $2^{4044}$, but this is a contradiction since $p-2^{2022}>2^{4044}$. It really makes no difference, but I'll add this because the $2^{4044}$ feels a bit strange for me. You only need that $p>2^{2022}+1$. Then $p-2^{2022}>1$ is odd (hence doesn't divide $2^{2022}$) and less than $p$ (hence doesn't divide $p$), therefore, it doesn't divide $2^{2022}p$.