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 $A_k=\{b_1{a}_1+\cdots +b_m{a}_m\mid b_1,\dots ,b_m=\pm 1,1\le m\le 2k\}$ 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 $A_k$. $\square$