Argentina_2018

Consider the sequence $\alpha$ with $17$ terms equal to $98$ and $2$ terms equal to $96$. Its sum is $17\cdot 98+2\cdot 96=1858>1810$, so $\alpha$ is admissible. Note that $\alpha$ has exactly two subsequence sums in the interval $[1810-48,1810+48]=[1762,1858]$. They are its extremes: $1858$ the sum of the entire sequence and $1762$, the sum of all terms except one $96$. This example shows that the minimum $d$ in question is at least $48$.

We show that each admissible sequence has a subsequence sum in the interval $[1762,1858]$, implying that the answer is $d_{\min }=48$. Suppose on the contrary that this is false for an admissible sequence $\beta$. Still more is it false for any subsequence of $\beta$. So by possibly removing terms one may assume that $\beta$ is minimal, with sum $S>1810$ but with sum $\le 1810$ of each proper subsequence. In fact the assumption then implies $S\ge 1859$ and $T\le 1761$ for every proper subsequence sum $T$. In particular, if $t$ is any term of $\beta$ then $S-t\le 1761$. Hence the inequalities $S\ge 1859$ and $S-t\le 1761$ imply $t\ge 1859-1761=98$. Each admissible sequence has at least $19$ terms (having sum $>1810$ and terms $\le 100$). Therefore $S\ge 98\cdot 19=1862$.

On the other hand, we proved the inequality $S-t\le 1761$ for any term $t$. Since $t\le 100$ by hypothesis, it follows that $S\le 1761+100=1861$, which yields a contradiction.