$n=2,3,4$ all work. $n$ cannot be negative or $0$. We check the rest of the positives. Answer is $9$.

We notice that $2^n$ grows very fast compared to n, so we only need to consider small n. Then, we can simply guess and check to find $n = 2,3,4$ so the answer is $9$

$n=2,3,4$ all work. $n$ cannot be negative or $0$. We check the rest of the positives. Answer is $9$.

If $n\geq 7$, then $RHS$ is positive, but $LHS$ is negative, so no solutions. If $n<0$, then $RHS$ is negative, but $LHS$ is positive, so no solutions. Hence, we need to check only cases where $0\leq n \leq 6$. $n=0$ and $n=6$ are obviously do not work, since one side is in that case $0$, but the other one is positive. $n=1$: $LHS=2\cdot 5=10$ and $RHS=8$, no solutions. $n=2$: $LHS=2^2 \cdot 4=16=8\cdot 2=RHS$. $n=3$: $LHS=8\cdot 3=RHS$ $n=4$: $LHS=16\cdot 2=32=8\cdot 4=RHS$ $n=5$: $LHS=2^5\cdot 1=32$ and $RHS=8\cdot 5=40$, so no solution. So the sum is $2+3+4=9$. Answer. $\boxed{9}.$

\[2^n(6-n)=8n.\]Notice $n$ can't be equal to $0$ $2^{3-n}=\frac{6}{n}-1$ $2^{3-n}=\frac{6}{n}-1 > 0$ Now, if $n < 0$, the given condition will never be fulfilled. Hence we need to focus on $n > 0$ $\frac{6}{n}>1 \implies 6>n$ We need to check for $n \in \{1,2,3,4,5\}$ Only $n=2,3,4$ satisfy the equation. the Answer is $\boxed{9}$