jmc

We can apply the Euclidean algorithm here. $$\begin{array}{rl}\gcd (3n+4,n)& =\gcd (n,3n+4-3n)\\ & =\gcd (n,4).\end{array}$$There are three cases to consider:

Case 1: $n$ is odd. Therefore, $n$ and 4 are relatively prime and have a greatest common divisor of 1.

Case 2: $n$ is a multiple of 2, but not a multiple of 4. In this case, $n$ and 4 share a common factor of 2. Since 4 has no other factors, $n$ and 4 have a greatest common divisor of 2.

Case 3: $n$ is a multiple of 4. In this case, $n$ and 4 have a greatest common divisor of 4.

Therefore, the three possible values for the greatest common divisor of $3n+4$ and $n$ are 1, 2, and 4. It follows that the sum of all possible value of the greatest common divisor of $3n+4$ and $n$ is $1+2+4=\boxed{7}$.