Let $n$ be a positive integer. An $n\times n$ board consisting of $n^2$ cells, each being a unit square colored either black or white, is called [i]convex[/i] if for every black colored cell, both the cell directly to the left of it and the cell directly above it are also colored black. We define the [i]beauty[/i] of a board as the number of pairs of its cells $(u,v)$ such that $u$ is black, $v$ is white, and $u$ and $v$ are in the same row or column. Determine the maximum possible beauty of a convex $n\times n$ board.

[i]Proposed by Ivan Novak[/i]
1. **Define the problem and notation:**
   We are given an \( n \times n \) board where each cell is either black or white. The board is called *convex* if for every black cell, the cell directly to the left and the cell directly above it are also black. We need to determine the maximum possible beauty of such a board, where beauty is defined as the number of pairs \((u, v)\) such that \(u\) is black, \(v\) is white, and \(u\) and \(v\) are in the same row or column.

2. **Define the number of black cells in each row:**
   Let \( a_i \) denote the number of black cells in the \( i \)-th row. Since the board is convex, \( a_i \geq a_{i+1} \) for all \( i \).

3. **Calculate the number of pr0 pairs in the same row:**
   The number of pr0 pairs with both cells in the same row is given by:
   \[
   \sum_{i=1}^{n} a_i (n - a_i)
   \]

4. **Calculate the number of pr0 pairs in the same column:**
   The number of columns with at least \( i \) black cells is at least \( a_i \). Hence, the number of columns with exactly \( i \) black cells is \( a_i - a_{i+1} \). Therefore, the number of pr0 pairs with both cells in the same column is:
   \[
   \sum_{i=1}^{n} i (n - i) (a_i - a_{i+1})
   \]

5. **Combine the two sums:**
   We want to maximize the total number of pr0 pairs:
   \[
   \mathcal{S} = \sum_{i=1}^{n} a_i (n - a_i) + \sum_{i=1}^{n} i (n - i) (a_i - a_{i+1})
   \]

6. **Simplify the expression:**
   Combining the sums, we get:
   \[
   \mathcal{S} = \sum_{i=1}^{n} a_i (n - a_i) + \sum_{i=1}^{n} i (n - i) (a_i - a_{i+1})
   \]
   Notice that the second sum can be rewritten using the telescoping property:
   \[
   \sum_{i=1}^{n} i (n - i) (a_i - a_{i+1}) = \sum_{i=1}^{n} a_i (2n + 1 - 2i - a_i)
   \]

7. **Maximize the function:**
   Consider the function \( f(a_i) = a_i (2n + 1 - 2i - a_i) \). The maximum value of \( f \) is attained at \( a_i = \{n - i, n + 1 - i\} \). Since \( a_i \geq a_{i+1} \), the board is convex.

8. **Calculate the desired sum:**
   The desired sum is:
   \[
   \sum_{i=1}^{n} (n - i)(n - i + 1) = \sum_{i=1}^{n} i^2 - i
   \]

9. **Sum the series:**
   The sum of the series is:
   \[
   \sum_{i=1}^{n} i^2 - \sum_{i=1}^{n} i = \frac{n(n+1)(2n+1)}{6} - \frac{n(n+1)}{2}
   \]
   Simplifying, we get:
   \[
   \frac{n(n+1)(2n+1)}{6} - \frac{n(n+1)}{2} = \frac{n(n+1)(2n+1) - 3n(n+1)}{6} = \frac{n(n+1)(2n+1 - 3)}{6} = \frac{n(n+1)(2n-2)}{6} = \frac{n(n+1)(n-1)}{3}
   \]

The final answer is \(\boxed{\frac{n^3 - n}{3}}\).