jmc

Let the given base $7$ number be $n$. Suppose that $n$ has $d+1$ digits in either base $7$ or base $16$. Let $a_d$ be the leftmost digit of $n$ in its base $7$ expression, $a_{d-1}$ be the digit that is second from the left, and so forth, so that $a_0$ is the base $7$ units digit of $n$. It follows that $a_d$ is the base $16$ units digit of $n$, and so forth. Converting to base $10$, it follows that $$n=7^d\cdot a_d+7^{d-1}\cdot a_{d-1}+\cdots +a_0=16^d\cdot a_0+16^{d-1}\cdot a_1+\cdots +a_d.$$Combining the like terms, it follows that $$(16^d-1)a_0+(16^{d-1}-7)a_1+\cdots +(1-7^d)a_d=0.$$For $d\le 3$, we observe that the powers of $16$ are significantly larger than the powers of $7$. More precisely, since $a_i\le 6$ for each $i$, then we have the following loose bound from the geometric series formula

$$\begin{array}{rl}0& =(16^d-1)a_0+(16^{d-1}-7)a_1+\cdots +(1-7^d)a_d\\ & \ge (16^d-1)+(1-7)\cdot 6+\cdots +(1-7^d)\cdot 6\\ & =16^d+d-6\cdot \frac{7^{d+1}-1}{7-1}\\ & \ge 16^d-7^{d+1}\end{array}$$For $d=3$, then $16^3=4096>7^4=2401$, and by induction, $16^d>7^{d+1}$ for all $d\ge 3$. Thus, $d\in \{0,1,2\}$. If $d=0$, then all values will work, namely $n=1,2,3,4,5,6$. If $d=1$, then $$(16-1)a_0+(1-7)a_1=15a_0-6a_1=3(5a_0-2a_1)=0.$$Thus, $5a_0=2a_1$, so $5$ divides into $a_1$. As $a_1\le 6$, then $a_1=0,5$, but the former yields that $n=0$. Thus, we discard it, giving us the number $n=52_7=5\cdot 7+2=37$. For $d=2$, we obtain that $$(256-1)a_0+(16-7)a_1+(1-49)a_2=3(51a_0+3a_1-16a_2)=0.$$Since $16a_2\le 6\cdot 16=96$, then $a_0=1$. Then, $51+3a_1=3(17+a_1)=16a_2$, so it follows that $a_2$ is divisible by $3$. Thus, $a_2=0,3,6$, but only $a_2=6$ is large enough. This yields that $a_1=15$, which is not possible in base $7$. Thus, the sum of the numbers satisfying the problem statement is equal to $1+2+3+4+5+6+37=\boxed{58}.$