Problem 1438

1. Initial Setup and Parity Analysis:
- John writes the numbers $1^2, 3^2, \ldots, (2n-1)^2$ on the blackboard. These numbers are all odd perfect squares.
- The sum of these numbers is initially even because the sum of an even number of odd numbers is even.

2. Transformation Step:
- John chooses three numbers $a_1, a_2, a_3$ from the blackboard, erases them, and writes the number $1 + \sum_{1 \leq i < j \leq 3} |a_i - a_j|$.
- We need to analyze the parity of the new number written on the board.

3. Parity of the New Number:
- The absolute difference $|a_i - a_j|$ between any two odd numbers is even because the difference between two odd numbers is even.
- Therefore, $\sum_{1 \leq i < j \leq 3} |a_i - a_j|$ is a sum of three even numbers, which is even.
- Adding 1 to this even sum results in an odd number: $1 + \text{even} = \text{odd}$.

4. Sum of Numbers on the Blackboard:
- Initially, the sum of all numbers on the blackboard is even.
- Each transformation step replaces three numbers with one odd number, reducing the total number of numbers by 2.
- The parity of the sum of the numbers on the blackboard remains even after each transformation because replacing three numbers with one odd number does not change the overall parity of the sum.

5. Final Two Numbers:
- Eventually, two numbers remain on the blackboard. Since the sum of all numbers on the blackboard is even, the sum of these two numbers must also be even.
- This implies that both remaining numbers are either both even or both odd.

6. Sum of Squares of the Final Two Numbers:
- If the final two numbers are both even, their squares are also even, and the sum of two even numbers is even.
- If the final two numbers are both odd, their squares are odd, and the sum of two odd numbers is even.
- Therefore, the sum of the squares of the final two numbers is always even.

7. Conclusion:
- The sum of the squares of the final two numbers is always even.
- The numbers $1^2, 3^2, \ldots, (2n-1)^2$ are all odd, and thus their sum is odd.
- Therefore, the sum of the squares of the final two numbers cannot be any of the numbers $1^2, 3^2, \ldots, (2n-1)^2$.

$\blacksquare$