Problem 10.6. Find the largest possible number $A$ having the following property: if the numbers $1,2, \ldots, 1000$ are ordered in arbitrary way then there exist 50 consecutive numbers with sum not less than $A$.

Ivan Landjev
10.6. Let $x_{1}, x_{2}, \ldots, x_{1000}$ be an arbitrary rearrangement of the numbers $1,2, \ldots, 1000$. Set

$$
S_{1}=x_{1}+x_{2}+\cdots+x_{50}, \ldots, S_{20}=x_{951}+x_{952}+\cdots+x_{1000}
$$

Since $S_{1}+\cdots+S_{20}=500500$, we have $S_{i} \geq \frac{500500}{20}=25025$ for at least one index $i$.

On the other hand, if a number $B$ has the required property then we have $B \leq 25025$. To see this consider the rearrangement

$$
1000,1,999,2, \ldots, 501,500
$$

and take arbitrarily fifty consecutive numbers in it. If the first number is greater than 500 , then the sum of these fifty numbers is 25025 , otherwise it is 25000 . Hence $A=25025$.