jmc

Obviously, $1$ is not a primitive root $(\mathrm{mod}7)$ (its powers are all congruent to $1$!).

Examining the powers of $2$, we see that $\{2^1,2^2,2^3,2^4,\dots \}\equiv \{2,4,1,2,\dots \}$ with repetition from this point onward. Since the powers of $2$ do not include all residues from $1$ to $6(\mathrm{mod}7)$, we see that $2$ is not a primitive root.

The logic of this example can be generalized. If $a$ is an integer and $a^k\equiv 1(\mathrm{mod}p)$, then the powers of $a$ repeat on a cycle of length at most $k$. Therefore, for $a$ to be a primitive root, it is necessary that $a^k\not\equiv 1(\mathrm{mod}p)$ for all positive $k$ smaller than $p-1$. Conversely, if $a^k\equiv 1(\mathrm{mod}p)$ for some positive $k$ smaller than $p-1$, then $a$ is not a primitive root $(\mathrm{mod}p)$. As examples, $4$ and $6$ are not primitive roots $(\mathrm{mod}7)$, because $4^3\equiv 1(\mathrm{mod}7)$ and $6^2\equiv 1(\mathrm{mod}7)$.

This leaves $3$ and $5$ as candidates. Checking the powers of $3$ and $5$ modulo $7$, we see that $$\begin{array}{r}{3}^1\equiv 3,3^2\equiv 2,3^3\equiv 6,3^4\equiv 4,3^5\equiv 5,3^6\equiv 1;\\ 5^1\equiv 5,5^2\equiv 4,5^3\equiv 6,5^4\equiv 2,5^5\equiv 3,5^6\equiv 1.\end{array}$$Thus, $3,5$ are primitive roots of $7$, so the desired sum is $3+5=\boxed{8}$.