China Mathematical Olympiad

For given $M > 2$, we construct by induction a sequence $\{a_n\}$ that satisfies the requirements. Take $a_1, a_2$ that satisfy $a_2 - a_1 = 1$ and $a_1 > M^2$. Now suppose $a_1, a_2, \dots, a_{2k}$ are already chosen, such that $a_i > M^i$, $i = 1, 2, \dots, 2k$ and such that the set $A_k = \{b_1a_1 + \dots + b_m a_m \mid b_1, \dots, b_m = \pm 1, 1 \le m \le 2k\}$ does not contain $0$. It is obvious that $A_k$ is symmetric, i.e., $A_k = -A_k$. $A_1 = \{a_1, -a_1, 1, -1\}$.

Let $n$ be the smallest positive integer not in $A_k$, $N = \sum_{i=1}^{2k} a_i$, now choose positive integers $a_{2k+1}, a_{2k+2}$ satisfying $a_{2k+2} - a_{2k+1} = N + n$, $a_{2k+1} > M^{2k+2}$, $a_{2k+1} > \sum_{i=1}^{2k} a_i$. We now show that $A_{k+1}$ does not contain $0$ and $n \in A_{k+1}$. First, $n = -\sum_{i=1}^{2k} a_i - a_{2k+1} + a_{2k+2}$.

On the other hand, if $\sum_{i=1}^m b_i a_i = 0$, $m \le 2k+2$, as $0 \notin A_k$, we must have $m = 2k+1$ or $2k+2$.

If $m = 2k+1$, then $\left| \sum_{i=1}^{2k+1} b_i a_i \right| \ge a_{2k+1} - \sum_{i=1}^{2k} a_i > 0$.

If $m = 2k + 2$ and $b_{2k+1}$ and $b_{2k+2}$ are of the same sign, then
$\left| \sum_{i=1}^{2k+2} b_i a_i \right| \ge a_{2k+1} + a_{2k+2} - \sum_{i=1}^{2k} a_i > 0$; if $b_{2k+1}$ and $b_{2k+2}$ are of different signs, then
$$\left| \sum_{i=1}^{2k+2} b_i a_i \right| = \left| \sum_{i=1}^{2k} b_i a_i \pm (a_{2k+1} - a_{2k+2}) \right| \ge | a_{2k+1} - a_{2k+2} | - \sum_{i=1}^{2k} a_i = N + n - N = n > 0.$$
The $\{a_n\}$ thus constructed satisfies the requirements since $0$ is not contained in any $A_k$, and any non-zero integer between $-k$ and $k$ is contained in $A_k$. $\square$