Suppose otherwise, that there exists an $n<22$ such that any $n$ consecutive integers contains a group of 7 that are pairwise coprime. Note that this implies that any group of 21 consecutive integers also contains a group of 7 that are pairwise coprime (simply consider a set of $n$ consecutive integers among the 21). Then consider a sequence of 21 consecutive integers, beginning on some number $a$ divisible by 2 and 5, congruent to 2 mod 3, congruent to 4 mod 7, 2 mod 11, and 2 mod 13. It can be verified that any number between $a$ and $a+20$ is divisible by at least 1 of $\{2,3,5,7,11,13\}$. Suppose we choose any 7 numbers in the range. By pigeonhole, at least one of the aforementioned primes divides two of these 7 numbers, hence proving the result by contradiction.