jmc

We can ignore the $0$ digit for now, and find the product of $24_7\times 3_7$. First, we need to multiply the units digit: $4_7\times 3_7=12_{10}=15_7$. Hence, we write down a $5$ and carry-over the $1$. Evaluating the next digit, we need to multiply $2_7\times 3_7+1_7=7_{10}=10_7$. Thus, the next digit is a $0$ and $1$ is carried over. Writing this out: \begin{array}{@{}c@{\;}c@{}c@{}c@{}c@{}c@{}c} & & & & & \stackrel{1}{2} & \stackrel{}{4}_7 \\ & & & \times & & & 3_7 \\ \cline{4-7} & & & & 1 & 0 & 5_7 \\ \end{array} We can ignore the $0$ in $30_7$, since it does not contribute to the sum. Thus, the answer is $1+0+5=\boxed{6}$.

Notice that the base seven sum of the digits of a number leaves the same remainder upon division by $6$ as the number itself.