3. In a $2 \times 2019$ grid, each cell is filled with a real number, where the 2019 real numbers in the top row are all different, and the bottom row is also filled with these 2019 real numbers, but in a different order from the top row. In each of the 2019 columns, two different real numbers are written, and the sum of the two numbers in each column is a rational number. How many irrational numbers can appear at most in the first row?
3. There can be at most 2016 irrational numbers.

It can be proven: In the table filled according to the rules, at least three rational numbers must appear in the first row (thus no more than 2016 irrational numbers).

In fact, each number will appear twice in the table, once in the upper row and once in the lower row.
First, observe a column of numbers.
Let the upper number be \(a_{1}\), and the lower number be \(a_{2}\). For convenience, denote this as \((a_{1}, a_{2})\), called a pillar. Color the pillar \((a_{1}, a_{2})\) red. Find another pillar where \(a_{2}\) is on top, and color that pillar red. If that pillar is \((a_{2}, a_{1})\), then stop; otherwise, that pillar is \((a_{2}, a_{3})\) \((a_{3} \neq a_{1})\). Continue coloring the pillar where \(a_{3}\) is on top, and so on, until \(a_{1}\) appears at the bottom of a pillar (this is feasible because for \(a_{2}\), \(a_{3}\), etc., they are colored in pairs). When the process ends, a chain of red pillars is obtained:
\[
\left(a_{1}, a_{2}\right), \left(a_{2}, a_{3}\right), \cdots, \left(a_{k}, a_{1}\right),
\]
which is called a "chain of length \(k\)." If there are still uncolored pillars, another chain can be found, and so on. Eventually, all pillars are divided into several disjoint chains. Since the sum of the lengths of all chains is 2019, there must be a chain of odd length.
Now observe the chain of odd length:
\[
\begin{array}{l}
\left(a_{1}, a_{2}\right), \left(a_{2}, a_{3}\right), \cdots, \left(a_{2t+1}, a_{1}\right). \\
\text{Let } a_{1} + a_{2} = b_{1}, a_{2} + a_{3} = b_{2}, \cdots, \\
a_{2t+1} + a_{1} = b_{2t+1},
\end{array}
\]
where \(t\) is a positive integer.
According to the problem, \(b_{i}\) is a rational number.
Thus,
\[
\begin{aligned}
2a_{1} &= \left(a_{1} + a_{2}\right) - \left(a_{2} + a_{3}\right) + \left(a_{3} + a_{4}\right) - \cdots - \\
& \left(a_{2t} + a_{2t+1}\right) + \left(a_{2t+1} + a_{1}\right) \\
&= b_{1} - b_{2} + b_{3} - \cdots - b_{2t} + b_{2t+1}
\end{aligned}
\]
is a rational number.
Therefore, \(a_{1}\) is a rational number.
Similarly, \(a_{1}, a_{2}, \cdots, a_{2t+1}\) are all rational numbers.
There are \(2t + 1 \geq 3\) such rational numbers.
Table 1 is an example filled according to the rules, where 2016 irrational numbers appear in the first row.
Table 1
\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|}
\hline 1 & 2 & 3 & $1+\sqrt{2}$ & $1-\sqrt{2}$ & $2+\sqrt{2}$ & $2-\sqrt{2}$ & $\cdots$ & $1008+\sqrt{2}$ & $1008-\sqrt{2}$ \\
\hline 2 & 3 & 1 & $1-\sqrt{2}$ & $1+\sqrt{2}$ & $2-\sqrt{2}$ & $2+\sqrt{2}$ & $\cdots$ & $1008-\sqrt{2}$ & $1008+\sqrt{2}$ \\
\hline
\end{tabular}