Let $ A$ be the arbitrary set of $ 16$ consecutive integers. It is clear that if $ p$ is a common prime factor of two elements of $ A$ then $ p \in \{2, 3, 5, 7, 11, 13 \}$. Thus, it is enough to show that we can find an element of $ A$ which is not divisible by any number from this set. Consider the following residues mod $ 30$: $ 1, 7, 11, 13, 17, 19, 23, 29$. They are coprime with $ 30$ and note that no matter how we choose the $ 16$ consecutive residues there are always $ 4$ of them belonging to this set. The difference between two of this numbers is never divisible by $ 7$ and $ 13$ and the only differences divisible by $ 11$ are $ 23 - 1 = 22 = 29 - 7$ but this numbers are too far of each other to be in the same set of $ 16$ consecutive integers. Hence among those $ 4$ numbers there is at most one number divisible by $ 7$, at most one number divisible by $ 11$ and at most one number divisible by $ 13$, so we can choose a number which is not divisible by any of them and since this number is also coprime to $ 30$ we are done.

I have a question about this solution. What if the first, second or third number in $ A$ is congruent to 17, 23 or 29 mod 30, and is also divisible by 7? Then the 14th, 15th or 16th is also, but is relatively prime to 30, as is asserted is impossible by the previous post. Here is a proposed fix: If the 1st, 2nd, 3rd number is 17 or 29 mod 30 and divisible by 7, then there are 3 other residues in $ A$ relatively prime to 30. At most one is divisible by 11, and at most one is divisible by 13, so there is one not divisible by any prime less than 16 as sought. Now if the 1st, 2nd, 3rd is 23 mod 30 and divisible by 7, then 29 and 1 are also in $ A$. These are both relatively prime to every number in the set even if one of them is divisible by 11 or 13, because the nearest multiples of 11 or 13 are outside of $ A$, as 11 from 1 mod 30 is 20 mod 30 which is too far to be inside the bounds. I'm sorry if I'm not clear, but there was a gaping hole in TomciO's proof I just had to fix.