The main observation is that if $s(a) = s(b)$, then $a = b$, as the problem condition insists on $a$ dividing $b$ and $b$ dividing $a$. Everything else is grinding. Note that $s(a) \leq 18$ for all $a\in A$ and hence $A$ can contain at most $18$ elements, one of each digit sum. Suppose, for contradiction, that $A$ with $18$ elements exists. Then each digit sum appears exactly once. The only number with sum $18$ is $99$, so it must be in $A$. The only numbers with sum $2$ are $2$, $11$, $20$, but only $11$ divides $99$, so $11$ must be in $A$. Then from sum $10$ we can have (by divisibility with $11$) $55$, then from sum $5$ we can have (as it needs to be a divisor of $55$) $5$. But now there is no number with sum $15$ which is divisible by $5$ - hence sum $15$ is actually not attainable, contradiction. So $A$ can have at most $17$ elements and we saw that we have to omit sum $15$ if we keep sum $18$. Now by continuing in the above fashion (determining the unique possible integer for each sum), we arrive at the following example: $1$, $11$, $3$, $22$, $5$, $33$, $7$, $44$, $9$, $55$, $92$, $66$, $94$, $77$, (no number with sum $15$), $88$, $89$, $99$.

Just a little correction, we don't have to omit sum 15 in order to keep sum 18, it's just that we can't have numbers with sums 2, 5, 10 and 18 all at the same time. An example with all sums except 2 would be: 1, 3, 22, 32, 33, 43, 44, 9, 64, 47, 66, 67, 86, 96, 88, 98, 99.

$\boxed{ANSWER:K_{max}=17}$ claim:$19>k$ proof:let's say $s(a)=s(b)\implies s(a)|s(b),s(b)|s(a)\implies a|b,b|a\implies a=b\implies k_{max}<19$ since we have $18$ different sum of digits in $A$ claim:$k\neq 18$ proof:let's say $k=18$ if $a\in A$ $s(a)=18\implies a=99\implies 99\in A$ $b\in A$ and $s(b)=2\implies b=2,11,20$ since $b,a\in A$ and $s(b)|s(a)\implies b|a\implies b=11$ $c\in A$ and $s(c)=10\implies c=91,19,28,82,73,37,64,46,55$ since $b,c\in A$ and $s(b)|s(c)\implies b|c\implies c=55$ $d\in A$ and $s(d)=5\implies d=5,50,14,41,23,32$ since $c,d\in A$ and $s(d)|s(c)\implies d|c\implies d=5$ $e\in A$ and $s(e)=15\implies e=96,69,87,78$ since $d,e\in A$ and $s(d)|s(e)\implies d|e$ which is condratiction because $5\nmid 96,69,87,78\implies k=17$ construction:

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