11.3. Find the maximum natural number $A$ such that for any arrangement of all natural numbers from 1 to 100 inclusive in a row in some order, there will always be ten consecutively placed numbers whose sum is not less than $A$.
Answer: 505.

Solution: The sum of all numbers from 1 to 100 is 5050. Let's divide the 100 numbers in a row into 10 segments, each containing 10 numbers. Clearly, the sum of the numbers in one of these segments is not less than 505, so A is not less than 505.

We will show that among the numbers arranged in the following order: $100,1,99,2,98,3, \ldots, 51,50$, there are no ten consecutive numbers whose sum is greater than 505. This will prove that A is not greater than 505, and considering the above, that $A=505$.

We will use the fact that in this arrangement, all numbers in odd positions decrease monotonically, while those in even positions increase monotonically, and the sum of a number in an odd position and the one following it is always 101. Therefore, the sum of any 10 consecutive numbers, the first of which is in an odd position, is always $505=5 * 101$. The sum of a segment of 10 consecutive numbers, the first of which is in an even position, is 505 minus the rightmost number among those to the left of this segment plus the last number of this segment. The two mentioned numbers are in odd positions, differing by 10, so the first is 5 greater than the second, and thus the entire sum is 500, which is less than 505.

Grading Criteria: Proof that A is not less than 505: 3 points. No justification for the lower bound $\mathrm{A}=505: -2$ points or -1 point.

Example proving that A is not greater than 505: 4 points.

Insufficient justification of the example: -2 points.