Answer: $2n^2 - n$.
Consider the position of the painting for each person in their preference order, and let $S$ denote the sum of these positions. The maximum of $S$ is $2n \cdot 2n = 4n^2$ and the minimum of $S$ is $2n \cdot 1 = 2n$. For each trade, $S$ is reduced at least by 2. Hence the maximum number of trades is $n(2n - 1)$.

Now we show that $n(2n - 1)$ trades can happen. We number the paintings from 1 to $2n$ in clockwise direction. Then we let odd-numbered paintings go clockwise and even-numbered paintings go counter-clockwise. After $n$ trades, we can have all odd-numbered paintings shifted one position in clockwise direction and all even-numbered paintings shifted one position in counter-clockwise direction. Repeating this $2n$ times will return the paintings to the initial position. Repeating $2n - 1$ times gives $n(2n - 1)$ trades. This determines the preference order for each person.