Blank Space Part 6

Let the sum be $N = 1 \square 2 \square 3 \square 4 \square 5 \square 6$ . Consider all the squares be switches and the switch before $2$ be $s_2$ , $3$ be $s_3$ , $4$ be $s_4$ , $5$ be $s_5$ and $6$ be $s_6$ . The maximum value of $N$ is $21$ when all the switches are $\color{#20A900}{\text{ON}}$ or $\color{#20A900} {+}$ . We note that when we switch $s_n$ $\color{#D61F06}{\text{OFF}}$ or to $\color{#D61F06}{-}$ , we minus $2n$ from $21$ . For $N = 7 = 21 -14 = 21 - 2(7)$ . We need to switch $\color{#D61F06}{\text{OFF}}$ switches, where the sum of their numbers is $7$ . And there are only $\boxed{2}$ ways to do this; that is when $3+4$ and $2+5$ .