Oleg drew an empty 50×50 table and wrote a number above each column and to the left of each row.

It turned out that all 100 written numbers are distinct, with 50 of them being rational and the other 50 irrational. Then, in each cell of the table, he wrote the sum of the numbers written next to its row and its column ("addition table"). What is the maximum number of sums in this table that could be rational numbers?
Evaluation. Let along the left side of the table be written \( x \) irrational and

\( 50-x \) rational numbers. Then along the top side, \( 50-x \) irrational and \( x \) rational numbers are written. Since the sum of a rational and an irrational number is always irrational, in the table there are at least \( x^2 + (50-x)^2 = 2(x-25)^2 + 2 \cdot 25^2 \geq 2 \cdot 25^2 = 1250 \) irrational numbers. Therefore, there are no more than \( 2500 - 1250 = 1250 \) rational numbers.

Example, when there are exactly 1250 rational numbers in the table. Place along the left side the numbers \( 1, 2, \ldots, 24, 25, 1+\sqrt{2}, 2+\sqrt{2}, \ldots, 25 \pm \sqrt{2} \), and along the top - the numbers \( 26, 27, \ldots, 49, 50 \),

\( 26-\sqrt{2}, 27-\sqrt{2}, \ldots, 50-\sqrt{2} \). Then only \( 2 \cdot 25^2 = 1250 \) sums of a rational and an irrational number will be irrational.

## Answer

1250 sums.