let $S_{m}=\sum\limits_{i=1}^m a_i-\sum\limits_{i=m+1}^n a_i $ and $S_{0}=-S_{n}$ so $S_{m}-2a_{m}=S_{m-1}$.now look at the sequence $S_{n},S_{n-1},...,S_{1},S_{0}$. we can assume that $S_{n}$ is not zero.suppose that $S_{n}>0$. we can find $i\geq 1$ such that $S_{i}\geq 0\geq S_{i-1}$.assume the contrary then $ | S_{i} |> | a_{i} |$ so $S_{i}> | a_{i} |$. and also $ | S_{i-1} |> | a_{i} |$ so $-S_{i-1}> | a_{i} |$ but $S_{i}-2a_{i}=S_{i-1}$ so $-S_{i}+2a_{i}> | a_{i} |$ and hence $2a_{i}>S_{i}+ | a_{i} |>2 | a_{i} |$ and then we get the contradiction...

Let $A=\sum\limits_{i=1}^n a_i$ and we have an $a_{k}$ where $k\leq n$ and $ |a_{k}|\geq |a_{i}|$ , $i=1, 2, ..., n$ , we will show that we can find an m for $a_{k}$, So the given inequality holds if and only if $\dfrac {A- |a_{k}|}{2}\leq \sum\limits_{i=1}^m a_{i}\leq \dfrac {A+ |a_{k}|}{2}$ so now let t be the greatest integer that satisfies $\dfrac {A- |a_{k}|}{2}>\sum\limits_{i=1}^t a_{i}$ if we do not have such m then $\dfrac{A+ |a_{k}|}{2} < \sum\limits_{i=1}^{t+1} a_{i}$ if we combine the last two inequalities we have $a_{t+1}> |a_{k}|$ , contradiction. So m=t+1 satisfies the problem.

Poland 1974 final test Prove that for all $ n\in Z_+ $ and any sequence of real numbers $ a_1, a_2, \ldots, a_n $there exists $ k \leq n \wedge k\in Z_+$ such that $$|\sum_{i=1}^k a_i-\sum_{i=k+1}^na_i|\le \max\lbrace |a_1|,|a_2|,...,|a_n|\rbrace$$

WolfusA wrote: Poland 1974 final test Prove that for all $ n\in Z_+ $ and any sequence of real numbers $ a_1, a_2, \ldots, a_n $there exists $ k \leq n \wedge k\in Z_+$ such that $$|\sum_{i=1}^k a_i-\sum_{i=k+1}^na_i|\le \max\lbrace |a_1|,|a_2|,...,|a_n|\rbrace$$ Does it have solution?

raven01 wrote: Let $A=\sum\limits_{i=1}^n a_i$ and we have an $a_{k}$ where $k\leq n$ and $ |a_{k}|\geq |a_{i}|$ , $i=1, 2, ..., n$ , we will show that we can find an m for $a_{k}$, So the given inequality holds if and only if $\dfrac {A- |a_{k}|}{2}\leq \sum\limits_{i=1}^m a_{i}\leq \dfrac {A+ |a_{k}|}{2}$ so now let t be the greatest integer that satisfies $\dfrac {A- |a_{k}|}{2}>\sum\limits_{i=1}^t a_{i}$ if we do not have such m then $\dfrac{A+ |a_{k}|}{2} < \sum\limits_{i=1}^{t+1} a_{i}$ if we combine the last two inequalities we have $a_{t+1}> |a_{k}|$ , contradiction. So m=t+1 satisfies the problem. You should explain why we can let t be the greatest integer that satisfies $\dfrac {A- |a_{k}|}{2}>\sum\limits_{i=1}^t a_{i}$