Let $a_1, a_2, \dots, a_n$ be the numbers in a line on the blackboard. Since $a_1 + a_2 + \dots + a_{40} = 0$, it follows that half of the numbers $a_1, a_2, \dots, a_{40}$ are equal to $1$ and the other half are equal to $-1$. Since $a_1 + a_2 + \dots + a_{40} + a_{41} + a_{42} \neq 0$ it follows that $a_{41} = a_{42} = a \in \{-1, 1\}$.

Similarly, $a_2 + a_3 + \dots + a_{43} \neq 0$ and hence $a_{42} = a_{43} = a \in \{-1, 1\}$. Proceeding in the same way we conclude that all numbers after $a_{41}$ must be equal to $a$.

Therefore, if $n > 60$, the sum $a_{22} + a_{23} + \dots + a_{61}$ has $a_{41} = a_{42} = \dots = a_{61} = a$, and hence $1$ and $-1$ cannot have the same cardinality in the set. It means that $a_{22} + a_{23} + \dots + a_{61} \neq 0$, and that contradicts condition (i).

Hence the greatest value of $n$ is $60$.

The maximal value of $\Sigma_n$ is obtained when we have as many as possible numbers equal to $1$. We remind that between the first $40$ numbers we must have $20$ numbers equal to $-1$. Hence the numbers equal to $1$ can be at most $40$. Therefore $\Sigma_n \le 20$. In fact, we observe that $\max \Sigma_n = 20$. It happens for
$$a_1 = \dots = a_{20} = 1,\ a_{21} = \dots = a_{40} = -1 \text{ and } a_{41} = \dots = a_{60} = 1$$