Lemma: Let $X$ and $Y$ be painted points on $AB$. If $XY=AB\cdot\frac{c}{d}$, $ \ gcd(c, d)=1$, and $p\mid d$ a prime divisor of $d$, then $p\mid u+1$, for some $u\in S$ Proof: We'll prove it using induction. The base case is trivial. We perform a move on some segment $XY$ dividing it into $n+1$ parts. Now for every segment $l$ it's lenght can expressed as $ZW+XY\cdot\frac{k}{n+1}$, where $XY\cdot\frac{e}{n+1}=l\cap XY$. But, $ZW+XY\cdot\frac{k}{n+1}=AB\cdot\frac{e}{f}+AB\cdot\frac{c}{d}\cdot\frac{k}{n+1}=AB\cdot\frac{e\cdot d\cdot (n+1)+c\cdot k\cdot f}{f\cdot d\cdot (n+1)}$. We just have to check the statement for $f\cdot d\cdot (n+1)$, but the statement is true for all the prime factors of $f\cdot d$ by the hypothesis, and it's trivial that it's true for $n+1$, completing the proof.$_{\blacksquare}$ Now, $\frac{AK}{KB}=\frac{2709}{2022}\rightarrow\frac{AK+KB}{KB}=\frac{2709+2022}{2022}\rightarrow KB=AB\cdot\frac{2022}{4731}=AB\cdot\frac{674}{19\cdot 83}$, thus $m\geq 82$, by the Lemma. Notice that $S$ must contain some $a\in\{18, 37, 56, 75\}$ and $b\in\{82\}$, because, then $19\mid a+1$ and $83\mid b+1$. In that case, just divide $AB$ in $(a+1)(b+1)$ parts. We just have to count the amount of subsets of $\{1, 2,\dots, 81\}$ that contain some element of $\{18, 37, 56, 75\}$. The total amount of subsets is $2^{81}$, and the amount that does not contain any element of $\{18, 37, 56, 75\}$ is $2^{77}$, thus the answer is $m=82$ and the number of subsets $S$ of $\{1, 2,\dots, 82\}$ that allows us to paint $K$ is $2^{81}-2^{77}$.$_{\blacksquare}$

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