1992
Problem - 177
Find that largest integer $A$ that satisfies the following property: in any permutation of the sequence $1001$, $1002$, $1003$, $\cdots$, $2000$, it is always possible to find $10$ consecutive terms whose sum is no less than $A$.
The answer is $15005$.
For all the permutations, let's cut the $1000$ terms equally into $100$ segments each of which contains $10$ terms. It is easy to see that at least one segment has a sum no less that $$\dfrac{1}{100}\displaystyle\sum_{k=1001}^{2000}k=15005$$