The optimal value of $r$ is $\frac{1}{252}$. This is attained by letting $S$ be the set of integers $n$ for which $\nu_{2}(n) \equiv 4 \bmod 5$ and $\nu_{3}(n) \equiv 1 \bmod 2$. Let $S$ be a set of positive integers satisfying the two conditions. For each prime $p$, let $A_{p}=\left\{\nu_{p}(n)\right.$ : $n \in S\}$. We claim that in fact $S$ is precisely the set of positive integers $n$ for which $\nu_{p}(n) \in A_{p}$ for each prime $p$. Let $p$ be prime and suppose that $a_{1} p^{e_{1}}, a_{2} p^{e_{2}} \in S$, with $p \nmid a_{1}, a_{2}$. Then, setting $b_{1}=p^{e_{1}}$ and $b_{2}=p^{e_{2}}$ in the second condition gives that $a_{1} p^{e_{2}} \in S$ as well. So, if we have an integer $n$ for which $\nu_{p}(n) \in A_{p}$ for each prime $p$, we can start with any element $n^{\prime}$ of $S$ and apply this step for each prime divisor of $n$ and $n^{\prime}$ to obtain $n \in S$. Now we deal with the first condition. Let $n$ be any positive integer. We will compute the least positive integer $m$ such that $m n \in S$. By the above result, we can work with each prime separately. For a given prime $p$, let $e_{p}$ be the least element of $A_{p}$ with $e_{p} \geq \nu_{p}(n)$. Then we must have $\nu_{p}(m) \geq e_{p}-\nu_{p}(n)$, and equality for all primes $p$ is sufficient. So, if the elements of $A_{p}$ are $c_{p, 1}<c_{p, 2}<c_{p, 3}<c_{p, 4}<\ldots$, then $$c_{p}=\max \left(c_{p, 1}, c_{p, 2}-c_{p, 1}-1, c_{p, 3}-c_{p, 2}-1, c_{p, 4}-c_{p, 3}-1, \ldots\right)$$ is the worst case value for $\nu_{p}(m)$. We conclude two things from this. First, we must have $\prod_{p} p^{c_{p}} \leq 100$ by condition 1, and in fact this is sufficient. Second, since we only care about $c_{p}$ and would like to minimize $r$, the optimal choice for $A_{p}$ is an arithmetic progression with first term $c_{p}$ and common difference $c_{p}+1$. So we assume that each $A_{p}$ is of this form. Let $t=\prod_{p} p^{c_{p}}$. We now compute $r$. Note that $S$ is the set of integers $n$ such that for each prime $p$, $$n \equiv a p^{k\left(c_{p}+1\right)-1} \bmod p^{k\left(c_{p}+1\right)}$$ for some positive integers $a, k$ with $a<p$. This means that each prime $p$ contributes a factor of $$\frac{p-1}{p^{c_{p}+1}}+\frac{p-1}{p^{2 c_{p}+2}}+\frac{p-1}{p^{3 c_{p}+3}}+\cdots=\frac{p-1}{p^{c_{p}+1}-1}=\frac{1}{1+p+\cdots+p^{c_{p}}}$$ to the density of $S$. Multiplying over all primes $p$ gives $r=\frac{1}{\sigma(t)}$, where $\sigma(t)$ is the sum of divisors of $t$. So, it suffices to maximize $\sigma(t)$ for $t \leq 100$. By inspection, $t=96$ is optimal, giving $r=\frac{1}{252}$.