jmc

Let $d=\gcd (n+7,2n+1)$, so $d$ divides both $n+7$ and $2n+1$. Then $d$ divides $2(n+7)-(2n+1)=13$, so $d$ is at most 13.

If $n=6$, then $\gcd (n+7,2n+1)=\gcd (13,13)=13$, which shows that the value of 13 is attainable. Therefore, the greatest possible value of $\gcd (n+7,2n+1)$ is $\boxed{13}$.