Problem 1760

1. Given that $n > 2$ is an even positive integer and $a_1 < a_2 < \dots < a_n$ are real numbers such that $a_{k + 1} - a_k \leq 1$ for each $1 \leq k \leq n - 1$. We need to show that:
$$\prod_{(i, j) \in A} (a_j - a_i) > \prod_{(i, j) \in B} (a_j - a_i)$$
where $A$ is the set of ordered pairs $(i, j)$ with $1 \leq i < j \leq n$ such that $j - i$ is even, and $B$ is the set of ordered pairs $(i, j)$ with $1 \leq i < j \leq n$ such that $j - i$ is odd.

2. First, observe that the condition $a_{k + 1} - a_k \leq 1$ implies that the differences between consecutive terms are at most 1. This means that the sequence $a_1, a_2, \ldots, a_n$ is relatively "tight" in terms of spacing.

3. Consider the pairs in $A$ and $B$. For $(i, j) \in A$, $j - i$ is even, so $j = i + 2m$ for some integer $m \geq 1$. For $(i, j) \in B$, $j - i$ is odd, so $j = i + 2m + 1$ for some integer $m \geq 0$.

4. Let's analyze the products:
$$\prod_{(i, j) \in A} (a_j - a_i) \quad \text{and} \quad \prod_{(i, j) \in B} (a_j - a_i)$$

5. For $(i, j) \in A$, since $j - i$ is even, the differences $a_j - a_i$ will generally be larger because the indices are further apart. Specifically, $a_j - a_i$ will be the sum of an even number of differences $a_{k+1} - a_k$, each of which is at most 1.

6. For $(i, j) \in B$, since $j - i$ is odd, the differences $a_j - a_i$ will generally be smaller because the indices are closer together. Specifically, $a_j - a_i$ will be the sum of an odd number of differences $a_{k+1} - a_k$, each of which is at most 1.

7. To make this more concrete, consider the smallest possible values for $n$. For $n = 4$, we have:
$$A = \{(1, 3), (2, 4)\} \quad \text{and} \quad B = \{(1, 2), (2, 3), (3, 4)\}$$
Then:
$$\prod_{(i, j) \in A} (a_j - a_i) = (a_3 - a_1)(a_4 - a_2)$$
$$\prod_{(i, j) \in B} (a_j - a_i) = (a_2 - a_1)(a_3 - a_2)(a_4 - a_3)$$

8. Since $a_{k+1} - a_k \leq 1$, we have:
$$a_3 - a_1 \leq 2 \quad \text{and} \quad a_4 - a_2 \leq 2$$
Therefore:
$$(a_3 - a_1)(a_4 - a_2) \leq 4$$

9. On the other hand:
$$(a_2 - a_1)(a_3 - a_2)(a_4 - a_3) \leq 1 \cdot 1 \cdot 1 = 1$$

10. Thus, for $n = 4$, we have:
$$\prod_{(i, j) \in A} (a_j - a_i) \leq 4 \quad \text{and} \quad \prod_{(i, j) \in B} (a_j - a_i) \leq 1$$
which shows that:
$$\prod_{(i, j) \in A} (a_j - a_i) > \prod_{(i, j) \in B} (a_j - a_i)$$

11. This argument can be generalized for larger even $n$ by noting that the product of differences for pairs in $A$ will always involve larger differences (since they are sums of even numbers of terms each at most 1), while the product of differences for pairs in $B$ will involve smaller differences (since they are sums of odd numbers of terms each at most 1).

Therefore, we have shown that:
$$\prod_{(i, j) \in A} (a_j - a_i) > \prod_{(i, j) \in B} (a_j - a_i)$$
$\blacksquare$