VMO

Firstly, we prove that if $m$ satisfies the first condition, then $m$ has only one prime divisor. Assume that $m$ has at least 2 prime divisors, let $p$ be the smallest prime divisor of $m$ and $p_1,p_2,\dots ,p_k$ be the remaining prime divisors of $m$. We have $$\frac{\phi (m)}{m}=\left(1-\frac{1}{p}\right)\left(1-\frac{1}{p_1}\right)\dots \left(1-\frac{1}{p_k}\right)<1-\frac{1}{p}=\frac{p-1}{p}.$$ Hence, by choosing $n=p$ in i), one can get $$\frac{s(p,m)}{m-p}=\frac{\phi (m)-(p-1)}{m-p}<\frac{\phi (m)-\frac{\phi (m)}{m}\cdot p}{m-p}=\frac{\phi (m)}{m}=\frac{s(1,m)}{m},$$ which is a contradiction. Therefore, $m$ must be a power of a prime. Let $m=p^k$, note that $$2022^{p^k}+1\equiv 2022+1\equiv 2023\equiv 0(\mathrm{mod}p),$$ thus $p\mid 2023$ and $p\in \{7,17\}$. If $p=7$, using LTE, we have $$v_7(2022^{7k}+1)=v_7(2023)+v_7(7^k)=1+k\ge v_7(7^{2k})=2k.$$ From this, we conclude that $k=1$ and $m=7$. Similarly, for $p=17$, applying LTE, we also have $$v_{17}(2022^{17k}+1)=v_{17}(2023)+v_{17}(17^k)=2+k\ge v_{17}(17^{2k})=2k,$$ so $k\in \{1,2\}$ and $m\in \{17,289\}$. Therefore, $m=7,17,289$ are all desired numbers. $\square$