Solution:
The two numbers must be $16$ and $17$.

First, note that if a number $n$ is not a power of a prime, then it has two smaller factors $a$ and $b$ that are relatively prime and multiply to $n$. Thus, $n$ is the least common multiple of $a$ and $b$. Then, any multiple of both $a$ and $b$ must also be a multiple of $n$, so any multiple of all the numbers less than $n$ would have to be a multiple of $n$ (since two of the numbers less than $n$ are $a$ and $b$). Thus, for there to be some number that is a multiple of all the numbers less than $n$ but not $n$ itself, $n$ must be a prime power. Therefore, both consecutive numbers are prime powers.

Note also that if $n$ is at most $15$, then $2n \leq 30$. Thus, if we chose one of the consecutive numbers to be $n \leq 15$, the big number would have to be a multiple of $2n$, and therefore also of $n$. Thus, this cannot happen, so both numbers must be more than $15$. The only pair of consecutive numbers from $16$ to $30$ that are both prime powers is $16$ and $17$, so this must be the answer.