jmc

Obviously, we have that $k>3$, because otherwise two of the integers would be identical and not be relatively prime. Start by testing $k=4$. $6n+4$ and $6n+3$ are relatively prime because they are consecutive integers, but $6n+4$ and $6n+2$ are both even and are therefore not relatively prime. The next candidate to test is $k=5$. Firstly, we have that $$\begin{array}{rl}\gcd (6n+5,6n+3)& =\gcd (6n+3,(6n+5)-(6n+3))\\ & =\gcd (6n+3,2).\end{array}$$Since $6n+3$ is always odd, the two integers $6n+5$ and $6n+3$ are relatively prime. Secondly, $$\begin{array}{rl}\gcd (6n+5,6n+2)& =\gcd (6n+2,(6n+5)-(6n+2))\\ & =\gcd (6n+2,3).\end{array}$$Note that $6n+3$ is always divisible by 3, so $6n+2$ is never divisible by 3. As a result, we have that $6n+5$ and $6n+2$ are relatively prime. Finally, $$\begin{array}{rl}\gcd (6n+5,6n+1)& =\gcd (6n+1,(6n+5)-(6n+1))\\ & =\gcd (6n+1,4).\end{array}$$Note that $6n+1$ is always odd, so $6n+5$ and $6n+1$ are also relatively prime. Therefore, the smallest positive integer $k$ that permits $6n+k$ to be relatively prime with each of $6n+3$, $6n+2$, and $6n+1$ is $k=\boxed{5}$.