A plane has several seats on it, each with its own price, as shown below(attachment). $2n-2$ passengers wish to take this plane, but none of them wants to sit with any other passenger in the same column or row. The captain realize that, no matter how he arranges the passengers, the total money he can collect is the same. Proof this fact, and compute how much money the captain can collect.
1. **Restate the problem in a more abstract form**:
   We have a grid of seats with prices, and we need to place $2n-2$ passengers such that no two passengers are in the same row or column. We need to prove that the total money collected is the same regardless of the arrangement and compute this total.

2. **Transform the problem**:
   Consider the grid as an $n \times n$ matrix where each cell $(i, j)$ represents the price of the seat in the $i$-th row and $j$-th column. We need to show that the sum of the prices of the seats occupied by the passengers is invariant under any arrangement that satisfies the given constraints.

3. **Diagonal argument**:
   We can use the concept of the main diagonal of the matrix (from the bottom-left to the top-right). We will show that we can always rearrange the passengers such that they occupy the diagonal without changing the total sum.

4. **Operation definition**:
   Define an operation that swaps the positions of two passengers in the same column but different rows. Specifically, if we have passengers at $(a, b)$ and $(c, d)$, we can swap them to $(a, d)$ and $(c, b)$.

5. **Induction base case**:
   For the base case, consider the smallest grid where $n=2$. The passengers can only occupy the diagonal positions $(1,1)$ and $(2,2)$ or $(1,2)$ and $(2,1)$. In both cases, the sum of the prices is the same.

6. **Induction hypothesis**:
   Assume that for an $n \times n$ grid, we can always rearrange the passengers to occupy the diagonal without changing the total sum.

7. **Induction step**:
   For an $(n+1) \times (n+1)$ grid, assume we have already arranged the first $n$ columns such that the passengers occupy the diagonal. For the $(n+1)$-th column, if the passenger is already on the diagonal, we are done. If not, the passenger must be below the diagonal (since being above would contradict the arrangement of the first $n$ columns). We can then use the defined operation to swap this passenger with another passenger in the same row but different column, ensuring the passenger moves to the diagonal.

8. **Conclusion**:
   By induction, we can always rearrange the passengers to occupy the diagonal, and since the sum of the prices on the diagonal is invariant, the total sum collected is the same for any valid arrangement.

9. **Compute the total sum**:
   The sum of the prices on the diagonal of an $n \times n$ grid is given by the sum of the first $n$ natural numbers, which is $\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$. However, since we have $2n-2$ passengers, we need to adjust this sum accordingly.

   The total sum collected is:
   \[
   \sum_{i=1}^{n-1} i + \sum_{i=1}^{n-1} (n-i) = \sum_{i=1}^{n-1} i + \sum_{i=1}^{n-1} i = 2 \sum_{i=1}^{n-1} i = 2 \cdot \frac{(n-1)n}{2} = (n-1)n
   \]

The final answer is $\boxed{(n-1)n}$.