I have no background in number theory, but here are my thoughts on this problem.
Let's take $n$ to be this natural number. Consider the sequence $p(k) = nk + (n-1)$. Now, this is a polynomial in $k$, so it cannot generate infinitely many primes in a row. So there must be a $K$ such that $p(K)$ is not a prime.
It is clear that $p(K)$ and $p(K+1)$ are coprimes, for if any number divides both of them, then it must divide their difference, which is $n$. But this leads to a contradiction since $p(k) = n(k+1)-1$.
Now, the only part remaining to find such a $K$ such that $p(K+1)$ is not prime.
My idea here is to take $K=a(n-1)$. We can clearly see that $p(K)$ is divisible by $n-1$. What I want now is to choose $a$ such that $p(K+1)$ is composite. But notice that $p(K+1) = a(n-1)n + 2n-1$. So take $a=2n-1$. Then we have:
$$(n-1)(2n-1)n + n-1$$
$$(n-1)(2n-1)n + 2n-1$$
Both composite numbers and distant by $n$.