L567 wrote: The sum of $n > 2$ nonzero real numbers (not necessarily distinct) equals zero. For each of the $2^n - 1$ ways to choose one or more of these numbers, their sums are written in non-increasing order in a row. The first number in the row is $S$. Find the smallest possible value of the second number. Let $x_1\ge x_2\ge ...\ge x_m>0>x_{m+1}\ge x_{m+2}...\ge x_n$ be the $n$ numbers. Since all numbers are nonzero, $n>2$ and $\sum x_i=0$, we have $n>m\ge 1$ Let $S,T$ be the two first numbers in the row : $S=\sum_{i=1}^mx_i$ and $T=S-\min(|x_m|,|x_{m+1}|)$ $S=\sum_{i=1}^mx_i\ge mx_m$ and so $|x_m|\le\frac Sm$ $0=\sum_{i=1}^nx_i\le \sum_{i=1}^mx_i+(n-m)x_{m+1}$ and so $|x_{m+1}|\le \frac S{n-m}$ So $\min(|x_m|,|x_{m+1}|)\le \min(\frac Sm,\frac S{n-m})=\frac S{\max(m,n-m)}$ So $T\ge S-\frac S{\max(m,n-m)}$ and so, for smallest $T$, we are looking for smallest $\max(m,n-m)$, which is $\left\lceil\frac n2\right\rceil$ And this smallest value can indeed be reached for example with numbers : $\overbrace{\frac S{\left\lceil\frac n2\right\rceil},\cdots,\frac S{\left\lceil\frac n2\right\rceil}}^{\times\left\lceil\frac n2\right\rceil},\overbrace{-\frac S{\left\lfloor\frac n2\right\rfloor},\cdots,-\frac S{\left\lfloor\frac n2\right\rfloor}}^{\times\left\lfloor\frac n2\right\rfloor}$ Hence answer $\boxed{S-\frac S{\left\lceil\frac n2\right\rceil}}$

Answer is $S(1-\frac{1}{m})$, where $m=[\frac{(n + 1)}{2}]$. ($[m]$ is the floor function). WLOG assume $a_1 \geq a_2 \geq ... \geq a_{n}$ and let's say $S=a_1+..+a_k=|a_{k+1}+...+a_n|$. It is not hard to notice that the second number in the list is $T = max(s-a_k,s+a_{k+1})$. Let's try and find maximum value of $a_k$. Since $S=a_1+..+a_k\geq ka_k$, we can say that maximum value of $a_k$ is $\frac{S}{k}$. With same logic, minimum value of $a_{k+1}$ is $-\frac{S}{n-k}$. So minimum value of $T = S\cdot max(1-\frac{S}{k}, 1-\frac{S}{n-k})$. Since we want to maximum between two quantities be as small as possible, $k$ and $n-k$ should be as close as possible. Since k is natural number than $T$ takes minimum value when $k = m$. By putting $k=m$ we have $T=S(1-\frac{1}{m})$.

My solution from contest that was graded as a 5. I got no response on coordination.

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