IMO Hk TST

The greatest $M$ is $4$ if $n = 4$, and is $3$ if $n \ge 5$.

Let $S$ be the sum $\sum_{a \in A} |a|$.

For $n = 4$, by considering $A = \{-1, 0, 1, 2\}$, we need $M \le 4$. Let $a < b < c < d$ be the elements in $A$. From the condition, we must have $b - a, c - b, d - c \ge 1$. This implies
$$S = (|a| + |d|) + (|b| + |c|) \ge |d - a| + |c - b| \ge 3 + 1 = 4.$$
This shows the largest $M$ is $4$ when $n = 4$.

For $n \ge 5$, by considering $A = \{-1, 0, \varepsilon, 2\varepsilon, \dots, (n-4)\varepsilon, 1 + (n-4)\varepsilon, 1 + (n-3)\varepsilon\}$ where $\varepsilon > 0$, we need $M \le 3 + f(n)\varepsilon$ for some function $f$. When $\varepsilon$ tends to $0$, this yields $M \le 3$.

Firstly, for $n = 5$, let $a < b < c < d < e$ be the elements in $A$. Since $c-a > c-b$ and $e-c > d-c$, we must have $c-a \ge 1$ and $e-c \ge 1$ by considering the element $c$. Also, at least one of $c-b \ge 1$ and $d-c \ge 1$ must hold. WLOG assume $c-b \ge 1$. Then we have
$$S \ge (|a| + |e|) + (|b| + |c|) \ge |e-a| + |c-b| \ge 2 + 1 = 3.$$
This shows the largest $M$ is $3$ when $n = 5$.

Secondly, for $n \ge 6$, let $a$ be the element with the smallest absolute value. WLOG assume $a \ge 0$.

* If $0 \le a \le \frac{1}{2}$, then whenever $|x-a| \ge 1$, we must have $|x| \ge 1-a$. Since there are three such $x$'s, we have $S \ge 3(1-a) + (n-3)a \ge 3(1-a) + 3a = 3$.
* If $a > \frac{1}{2}$, then all elements have absolute value larger than $\frac{1}{2}$, and hence $S > \frac{n}{2} \ge 3$.

This shows the largest $M$ is $3$ when $n \ge 6$.