Fib

Note that $\begin{aligned} f_{x+2} & = f_{x}+f_{x+1} \\ \ \\ f_{x+3} & = f_{x+2}+f_{x+1} \\ & = f_x+2f_{x+1} \\ \ \\ f_{x+4} & = f_{x+3}+f_{x+2} \\ & = 2f_x+3f_{x+1} \\ \ \\ f_{x+5} & = f_{x+4}+f_{x+3} \\ & = 3f_x+5f_{x+1} \\ \ \\ f_{x+6} & = f_{x+5}+f_{x+4} \\ & = 5f_x+8f_{x+1} \\ \ \\ f_{x+7} & = f_{x+6}+f_{x+5} \\ & = 8f_x+13f_{x+1} \end{aligned}$ The sum of the eight numbers is $21f_n+33f_{x+1}$ . Since $f_{x+8}=13f_x+21f_{x+1}$ and $f_{x+9}=21f_x+34f_{x+1}$ , $f_{x+8} \ < \ 21f_x+33f_{x+1} \ < f_{x+9}$ So $f_x+f_{x+1}+f_{x+2}+f_{x+3}+f_{x+4}+f_{x+5}+f_{x+6}+f_{x+7}$ is between two consecutive fibonacci number.

Therefore $\boxed{\text{it is not possible}}$ .