for $1001+1002+1003+...+1010$ for every permutation $(a_{1},...,a_{1000})$ of $\{1001,1002,...,2000\}$, $\forall \ k\in; \ [|1,990|] ;\ \sum_{i=0}^{9}a_{k+i}\ge 1001+1002+...+1010$ but for $(1001+1002+1003+...+1010)+1$ it we take the permutation $(a_{1},..,a_{1000})$ with $a_{i}=1000+i$ we have $\sum_{i=1}^{10}a_{i}=1001+1002+...+2000< (1001+1002+1003+...+1010)+1$ then $A=1001+1002+1003+...+1010=1055$

N.T.TUAN wrote: Find the greatest positive integer $A$ with the following property: For every permutation of $\{1001,1002,...,2000\}$ , the sum of some ten consecutive terms is great than or equal to $A$. Let $t_{k}=a_{k}+a_{k+1}+...+a_{k+9}$. We have $\sum_{k=1}^{100}t_{10k-9}=500*3001=1500500$. Therefore exist k, suth that $t_{k}\ge 15005$ for all permutations. These is better inequalities, because exist permutation, when $a_{k}=1000+(k+1)/2$ if k is odd, and $a_{k}=2001-k/2$ if k is even, when for any k $t_{k}=15005$.

aviateurpilot wrote: for $1001+1002+1003+...+1010$ for every permutation $(a_{1},...,a_{1000})$ of $\{1001,1002,...,2000\}$, $\forall \ k\in; \ [|1,990|] ;\ \sum_{i=0}^{9}a_{k+i}\ge 1001+1002+...+1010$ but for $(1001+1002+1003+...+1010)+1$ it we take the permutation $(a_{1},..,a_{1000})$ with $a_{i}=1000+i$ we have $\sum_{i=1}^{10}a_{i}=1001+1002+...+2000< (1001+1002+1003+...+1010)+1$ then $A=1001+1002+1003+...+1010=1055$ You are false! ''SOME'' not ''ANY''