A dividing pair

Group the numbers by their largest odd divisor.

1. Write each number as $2^a\times m$ with $m$ odd; there are $15$ possible odd parts $m$, from $1$ to $29$.
2. Numbers with the same odd part form a chain such as $3$, $6$, $12$, $24$, in which the smaller always divides the larger.
3. So among any $16$ numbers, $2$ share an odd part and one divides the other.
4. The $15$ numbers from $16$ to $30$ contain no such pair, because $2\times 16=32$ is larger than $30$, so $15$ numbers are not enough.
5. So the answer is $16$.

Answer: $16$