XXII OM - III - Problem 5

Find the largest integer $ A $ such that for every permutation of the set of natural numbers not greater than 100, the sum of some 10 consecutive terms is at least $ A $.
The sum of all natural numbers not greater than $100$ is equal to $1 + 2 + \ldots + 100 = \frac{1 + 100}{2} \cdot 100 = 5050$. If $a_1, a_2, \ldots, a_{100}$ is some permutation of the set of natural numbers not greater than $100$ and the sum of any $10$ terms of this permutation is less than some number $B$, then in particular

By adding these inequalities side by side, we get that $a_1 + a_2 + \ldots + a_{100} = 1 + 2 + \ldots + 100 < 10B$, which means $505 < B$.
Thus, the number $A$ defined in the problem satisfies the inequality

On the other hand, consider the following permutation $a_1, a_2, \ldots, a_{100}$ of the set of natural numbers not greater than $100$

This permutation can be defined by the formulas:

We will prove that the sum of any $10$ consecutive terms of this permutation is not greater than $505$.
Indeed, if the first of the considered $10$ terms has an even number $2k$, then

If, however, the first of the considered terms has an odd number $2k + 1$, then

Thus, the sum of any $10$ consecutive terms of this permutation is not greater than $505$. Therefore, the number $A$ defined in the problem satisfies the inequality

From (1) and (2), it follows that $A = 505$.
Note 1. The problem can be generalized as follows: Find the largest integer $A$ such that for any permutation of the set of natural numbers not greater than an even number $n = 2t$, the sum of some $m = 2r$ (where $r$ is a divisor of $t$) consecutive terms is at least $A$.
By making minor changes in the solution provided above, consisting in replacing the number $100$ with $2t$ and the number $10$ with $2r$, it can be proved that $A = \frac{1}{2} m(n + 1)$.
Note 2. In the case where $m \leq n$ are any natural numbers, it is generally not true that every permutation of the set of natural numbers not greater than $n$ contains $m$ consecutive terms with a sum not less than $\frac{1}{2} m(n + 1)$. For example, for $n = 6$, $m = 4$, the permutation $6, 4, 1, 2, 3, 5$ does not contain four consecutive terms with a sum not less than $14$.