Answer: Clues 15 and 16 were incorrect. So the number is not divisible by 16 and it’s not divisible by 17.
Bonus Answer: $2^3\cdot 3^3\cdot 5^2\cdot 7\cdot 11\cdot 13\cdot 19\cdot 23\cdot 29\cdot 31=2,123,581,660,200$
Solution. First note that if a number is not divisible by $n$, then it is not divisible by any multiple of $n$. So if the mystery number were not divisible by 2, then it would not be divisible by $4, 6, 8,\ldots$. That would then imply that more than exactly two consecutive clues were incorrect. So the mystery number must be divisible by 2. The same argument applies for any of the numbers up to and including 15 (if the number weren’t divisible by 15, then it couldn’t be divisible by 30) and does not apply to any of the remaining numbers. For example, if the mystery number were not divisible by 16, then we would know that it is not divisible by $32, 48,\ldots$, which it is not claimed to be.
So at this point, we can say that it may be false that the mystery number is divisible by both 16 and 17. We will now show that these are the only possibilities among the numbers greater than 15.
At this point, we know the mystery number is divisible by $2, 3, 4, \ldots , 15$. We will now use the fact that if a number is divisible by $m$ and $n$, where $m$ and $n$ share no common factors (other than 1), then the number is also divisible by $mn$.
Observe that $$18=2\cdot 9,\ 20=4\cdot 5,\ 22=2\cdot 11,\ 24=3\cdot 8,\ 26=2\cdot 13,\ 28=4\cdot 7,\ \text{and}\ 30=2\cdot 15$$Based on our previous comment, this shows that the mystery must be divisible by 18, 20, 22, 24, 26, 28, and 30. Therefore, there are no other candidates for consecutive numbers that the mystery number is not divisible by, leaving only 16 and 17 (which corresponds to clues 15 and 16).
The smallest number satisfying the clues that are correct is the least common multiple of the numbers: $2, 3, 4, \ldots , 15, 18, 19, 20, \ldots 31$. The least common multiple of these numbers can be obtained by multiplying the largest powers of each prime that divides a number in this list. Certainly $2^3$ is the largest power of the prime 2 that divides any number in the list as $2^4=16$, which doesn’t divide any number in the list (recall 16 isn’t in the list). Similarly, $3^3$ is the largest power of the prime 3 that divides a number in the list (the number 27); $5^2$ is the largest power of 5 that divides a number in the list (the number 25); and then except for 17, the highest power of each of the remaining primes up to 31 that divides a number in the list is 1. Therefore, the least common multiple of the numbers in the list is $$2^3\cdot 3^3\cdot 5^2\cdot 7\cdot 11\cdot 13\cdot 19\cdot 23\cdot 29\cdot 31=2,123,581,660,200$$