Let's determine the largest of the $A$ numbers for which the following statement holds. No matter how we write down the first 100 positive integers, there will always be ten consecutive numbers in the sequence whose sum is at least $A$.
Let the first 100 positive integers be in some order $a_{1}, a_{2}, \ldots, a_{100}$. Their sum is

$$
a_{1}+a_{2}+\ldots+a_{100}=(1+100)+(2+99)+\ldots+(50+51)=50 \cdot 101=5050
$$

Therefore, if we divide the terms of the sum $a_{1}+a_{2}+\ldots+a_{100}$ into ten groups of ten, there will be a group where the sum of the numbers is at least $5050: 10=505$.

This means that no matter how we write the first 100 positive integers, there will always be ten consecutive numbers in the sequence whose sum is at least 505. The sought number is therefore at least 505.

If there is an order of the first 100 positive integers where the sum of any ten consecutive numbers is at most 505, then we are done; in this case, the sought number is exactly 505, because choosing a larger number $A$ would not satisfy the condition for the given order. The following order is exactly such:

$$
100,1,99,2, \ldots, 51,50
$$

Here, the numbers in the odd positions decrease from 100 to 51, and the numbers in the even positions increase from 1 to 50. The sum of any two consecutive numbers is either 101 or 100, so the sum of any ten consecutive numbers is 505 or 500, depending on whether the first of the ten numbers is in an odd or even position in the sequence.

The sought number is therefore 505.