10.3. The cells of a $2 \times 2019$ table must be filled with numbers (exactly one number in each cell) according to the following rules. In the top row, there should be 2019 real numbers, none of which are equal, and in the bottom row, there should be the same 2019 numbers, but in a different order. In each of the 2019 columns, there should be two different numbers, and the sum of these two numbers should be a rational number. What is the maximum number of irrational numbers that could be in the first row of the table?

(S. Kudrya)
Answer: 2016.

Solution. Estimation. We will prove that in the first row of the table, where numbers are arranged according to the rules, there are no fewer than three rational numbers (and, accordingly, no more than 2016 irrational numbers). Each number appearing in the table is written in exactly two cells, one of which is in the top row, and the other in the bottom row. Consider some column, let the number in its top cell be \(a_1\), and in the bottom cell \(a_2\) (we will briefly denote such a column as \((a_1, a_2)\)). We will color the column \((a_1, a_2)\). We will find the column where the number \(a_2\) is in the top cell and color it. If this column is \((a_2, a_1)\), we will end the process. Otherwise, if this column is \((a_2, a_3)\), where \(a_3 \neq a_1\), we will continue: we will color the column where the number \(a_3\) is in the top cell, and so on - until we reach the column where \(a_1\) is in the bottom cell (this will definitely happen, since the numbers equal to \(a_2, a_3, \ldots\) are colored in pairs). At the end of the process, we will get a set of colored columns \((a_1, a_2), (a_2, a_3), \ldots, (a_k, a_1)\), which we will call a cycle of length \(k\). If there are still uncolored columns, we will highlight another cycle, and so on. In the end, the set of all columns of the table will be divided into non-intersecting cycles. Since the sum of the lengths of all cycles is 2019, there will be a cycle of odd length.

Consider this cycle: \((a_1, a_2), (a_2, a_3), \ldots, (a_{2t+1}, a_1)\), where \(t \geq 1\). By the condition, \(a_1 + a_2 = b_1, a_2 + a_3 = b_2, \ldots, a_{2t+1} + a_1 = b_{2t+1}\), where all \(b_i\) are rational numbers. Then \(2a_1 = (a_1 + a_2) - (a_2 + a_3) + (a_3 + a_4) - \ldots - (a_{2t} + a_{2t+1}) + (a_{2t+1} + a_1) = b_1 - b_2 + b_3 - \ldots - b_{2t} + b_{2t+1}\) is a rational number, so \(a_1\) is rational. Similarly, all numbers \(a_1, a_2, \ldots, a_{2t+1}\) are rational, and there are no fewer than \(2t+1 \geq 3\).

Example. We will provide an example of a table filled according to the rules, in the top row of which there are 2016 irrational numbers:

\[
\begin{array}{|l|l|l|l|l|l|l|l|l|l|}
\hline 1 & 2 & 3 & 1+\sqrt{2} & 1-\sqrt{2} & 2+\sqrt{2} & 2-\sqrt{2} & \ldots & 1008+\sqrt{2} & 1008-\sqrt{2} \\
\hline 2 & 3 & 1 & 1-\sqrt{2} & 1+\sqrt{2} & 2-\sqrt{2} & 2+\sqrt{2} & \ldots & 1008-\sqrt{2} & 1008+\sqrt{2} \\
\hline
\end{array}
\]

Remark. Note that the condition of the odd length of the row of the table is significant. For an even length of the row, it is not difficult to construct examples of tables in which all numbers are irrational.

The condition that a number does not stand under itself is also important, otherwise a cycle of length 1 could appear, and the answer to the problem would be 2018.

Comment. The correct estimation (no more than 2016 irrational or, equivalently, no fewer than 3 rational numbers) is proven - 5 points.

The statement about the partition of the permutation into independent cycles is accepted without proof.

If the estimation is not proven, it is proven that in the columns forming a cycle of odd length, all numbers are rational - 2 points (out of 5 possible points for the estimation).

A correct example with 2016 irrational numbers is provided - 2 points.

Proof of the irrationality of square roots of natural numbers that are not perfect squares, cube roots of natural numbers that are not perfect cubes, etc., is not required.