See that any sum containing $x_1$ should be smaller than $x_2+x_3+x_4$. (Because there are exactly $10$ sums which contains $x_1$ and $s_{11}=x_2+x_3+x_4$ is the smallest sum that does not contain $x_1$) Now, consider the sums that does not contain $x_1$. There are exactly $10$ of them and $6$ of them contains $x_6$. $s_{15}=x_2+x_3+x_6$ is the smallest number among those $6$ numbers. Hence, any sum containing $x_6$ is greater than any sum not containing $x_6$. This gives us $x_2+x_3+x_6>x_3+x_4+x_5\Rightarrow x_2+x_6>x_4+x_5$. Clearly, we have $m\leq10$ and $s_{10}=x_1+x_5+x_6$. Since $x_1+x_2+x_6>x_1+x_4+x_5$, we need to have $s_9=x_1+x_4+x_6, s_8=x_1+x_3+x_6$ and $s_7=x_1+x_2+x_6$. Thus, $m=7$. Example for $m=7$. $(x_1,x_2,x_3,x_4,x_5,x_6)=(-10000,1,2,4,8,10000)$.