Problem 1395

To prove the given statement, we will break it down into two parts: the "if" part and the "only if" part.

### "If" Part:
We need to show that if $a_1 - a_{2n} = n$, then $\{a_2, a_4, \ldots, a_{2n}\} = \{1, 2, \ldots, n\}$.

1. Sum of Differences:
The difference $a_1 - a_{2n}$ is the sum of the descents minus the sum of the ascents. The set of differences of consecutive elements is $\{1, 2, \ldots, 2n-1\}$. Therefore, the sum of all differences is:
$$\sum_{i=1}^{2n-1} i = \frac{(2n-1) \cdot 2n}{2} = n(2n-1)$$

2. Sum of Descents and Ascents:
Let the sum of the descents be $S_d$ and the sum of the ascents be $S_a$. We have:
$$S_d + S_a = n(2n-1)$$
Given $a_1 - a_{2n} = n$, we can write:
$$S_d - S_a = n$$

3. **Solving for $S_d$ and $S_a$**:
Adding and subtracting the two equations:
$$2S_d = n(2n-1) + n \implies S_d = n^2$$
$$2S_a = n(2n-1) - n \implies S_a = n^2 - n$$

4. Distribution of Elements:
Let the descents be $(a_j, a_{j+1})$ for $j \in J$. The sum of descents is:
$$\sum_{j \in J} a_j - \sum_{j \in J} a_{j+1} = n^2$$
Each element in group $A = \{n+1, n+2, \ldots, 2n\}$ must appear as the left element of a descent and never as a right element of a descent. This implies that no two elements of group $A$ can be next to each other.

5. Odd Indices:
Since $a_1$ is in group $A$ and $a_{2n}$ is in group $B$, all elements in group $A$ must have odd indices. Therefore, $\{a_2, a_4, \ldots, a_{2n}\} = \{1, 2, \ldots, n\}$.

### "Only If" Part:
We need to show that if $\{a_2, a_4, \ldots, a_{2n}\} = \{1, 2, \ldots, n\}$, then $a_1 - a_{2n} = n$.

1. Odd Indices:
Since elements in group $A$ have odd indices, the sum of descents is $n^2$.

2. Sum of Ascents:
By the previous calculation, the sum of ascents is $n^2 - n$.

3. Difference Calculation:
Using the relationship $S_d - S_a = n$, we get:
$$a_1 - a_{2n} = n$$

Thus, we have shown both directions of the statement.

$\blacksquare$