Round 3

Answer: (i) $\max M = 1$, (ii) $\min m = -\frac{n-2}{n-1}$.

i. For $(a_1, a_2, \dots, a_n) = (0, \dots, 0, 1)$, we have $M = 1$.
Now we show $M \le 1$ holds always. Suppose, on the contrary, that for some nice sequence $a_1, a_2, \dots, a_n$, we have $M > 1$. Let $T = (a_1 + \dots + a_n) - m - M$, then we have $T + M \le 1$ by the niceness condition, hence $T \le 1 - M < 0$. Then $m \le \frac{T}{n-2} < 0$ and thus $m + T < 0$, which contradicts niceness. Thus the maximum possible value for $M$ is 1.

ii. If $a_1, a_2, \dots, a_n$ is nice, then so is $b_1, b_2, \dots, b_n$ for
$$b_1 = \frac{1}{n-1} - a_1, \quad b_2 = \frac{1}{n-1} - a_2, \quad \dots, \quad b_n = \frac{1}{n-1} - a_n.$$
From (i), we have $\frac{1}{n-1} - m \le 1$, thus $m \ge \frac{1}{n-1} - 1$. The minimum of $m$ is achieved on the sequence $\left( \frac{1}{n-1} - 1, \frac{1}{n-1}, \dots, \frac{1}{n-1} \right)$.