Let $S$ denote the set of three-digit numbers that have digit sum equal to $9$ and no digit equal to $0$. We will first find the cardinality of $S$. We start from the number $111$ and each element of $S$ can be obtained from $111$ by a string of $6A'$s (which means that we add $1$ to the current digit). Then for example $324$ can be obtained from $111$ by the string AAGAGAAA. There are in total
$$\frac{8!}{6! \cdot 2!} = 28$$
such words, so $S$ contains $28$ numbers. Now, from the conditions (3), (4) and (5), if $\overline{abc}$ is in $S$ then each of the other numbers of the form $\overline{**c}$ cannot be in $S$, neither $\overline{*b*}$ can be, nor $\overline{a**}$.

Since there are $a+b-2$ numbers of the first category, $a+c-2$ from the second and $b+c-2$ from the third one. In these three categories there are
$$(a+b-2)+(a+c-2)+(b+c-2)=2(a+b+c)-6=2 \cdot 9-6=12$$
distinct numbers that cannot be in $S$ if $abc$ is in $S$. So, if $S$ has $n$ numbers, then $12n$ are the forbidden ones that are in $S$, but each number from $S$ can be a forbidden number no more than three times, once for each of its digits, so
$$n + \frac{12n}{3} \le 28 \Leftrightarrow n \le \frac{28}{5},$$
and since $n$ is an integer, we get $n \le 5$. A possible example for $n=5$ is
$$S = \{144, 252, 315, 423, 531\}.$$