Given any $10$ consecutive positive integers , does there exist one integer which is relatively prime to the product of the rest ? Given any $10$ consecutive positive  integers , does there exist one integer which is relatively prime to the product of the rest ?
Notice that one integer $x$ being coprime to the product of the rest is equivalent to it being coprime to each other $y$ in the set, which is equivalent to it sharing no prime divisors. This gives immediately that:


  $x$ cannot be divisible by $2$, $3$, or $5$, since for those primes, either $x+p$ or $x-p$ would have to be in the interval, given the interval has length $10$.


Clearly, we also have, for any prime $p$ equal to at least $11$ that if $x$ is in the interval $x+p$ and $x-p$ are not (and nor is any other multiple of $p$). Thus, if $x$ is also not divisible by $7$, it must be coprime to all the other integers in the desired interval.

However, any interval of length $10$ has at least one integer in it which is not divisible by $2$, $3$, $5$, or $7$. In particular, notice that there will be:


$5$ integers divisible by $2$.
At most $2$ odd integers divisible $3$.
At most $1$ odd integer divisible by $5$.
At most $1$ odd integer divisible by $7$.


Implying at least one integer is not divisible by any of those primes and hence satisfies the desired condition.
I assume the integer has to be one of the 10. In which case I can think of cases in which there definitely is - if there is any prime in the sequence greater than our equal to 11 for example, or any integer which is a product of those primes (e.g. 143=13*11)

Whether this is true for all such sets of 10 consecutive numbers, I don't know