Let $n\geq 2$ be a positive integer. The sets $A_{1},A_{2},\ldots, A_{n}$ and $B_{1},B_{2},\ldots, B_{n}$ of positive integers are such that $A_{i}\cap B_{j}$ is non-empty $\forall i,j\in\{1,2,\ldots ,n\}$ and $A_{i}\cap A_{j}=\o$, $B_{i}\cap B_{j}=\o$ $\forall i\neq j\in \{1,2,\ldots, n\}$. We put the elements of each set in a descending order and calculate the differences between consecutive elements in this new order. Find the least possible value of the greatest of all such differences.
1. **Define the sets and elements:**
   Let \( x_1 < x_2 < \ldots < x_k \) be all the elements of \( \bigcup_{i=1}^n A_i \cup B_i \). Let \( d \) be the largest difference between consecutive elements of a set when arranged in descending order.

2. **Initial assumption:**
   If there exists an element \( x_i \) that belongs to exactly one of the sets, say \( A_j \), then since \( x_i \notin B_r \) for all \( 1 \leq r \leq n \), we can select an element \( x_a \in A_j \cap B_1 \) and add \( x_i \) to \( B_1 \). This operation does not increase \( d \) because \( x_a \in B_1 \) as well.

3. **Simplify the problem:**
   We can assume that for all \( i \), \( x_i \in A_{i_1} \cap B_{i_2} \) for some \( 1 \leq i_1, i_2 \leq n \). This means every element of the union of all sets is an element of one \( A \) set and one \( B \) set.

4. **Algorithm to reduce the problem:**
   - **Step 1:** If there exists some \( z \in \mathbb{Z}_{>0} \) such that \( z \in \bigcup_{i=1}^n A_i \cup B_i \), \( z+1 \notin \bigcup_{i=1}^n A_i \cup B_i \), and \( \exists c > 1 \) such that \( z+c \in \bigcup_{i=1}^n A_i \cup B_i \), then we can strictly decrease \( S = \sum_{1 \leq i,j \leq n} |x_i - x_j| \) without increasing \( d \).
   - **Step 2:** If \( |A_i \cap B_j| > 1 \) for some \( 1 \leq i, j \leq n \), then we can decrease \( k \) without increasing \( d \).

5. **Proof of the algorithm:**
   - For **Step 1**, replacing all \( x_i > z \) by \( x_i - c + 1 \) maintains \( A_i \cap A_j = \emptyset \) and \( B_i \cap B_j = \emptyset \). This operation either does not change \( |x_i - x_j| \) or decreases it, so \( d \) is not increased and \( S \) is strictly decreased.
   - For **Step 2**, if \( y > z \) and \( y, z \in A_i \cap B_j \), deleting \( y \) from all sets it is in and performing **Step 1** ensures \( A_i \cap B_j \neq \emptyset \).

6. **Termination of the algorithm:**
   The algorithm is finite because performing **Step 1** or **Step 2** strictly decreases \( S \), which is a non-negative integer.

7. **Final configuration:**
   When the algorithm finishes, \( x_{i+1} - x_i = 1 \) for all \( i \), and \( |A_i \cap B_j| = 1 \) for all \( 1 \leq i, j \leq n \). This implies \( |A_i| = |B_i| = n \) for all \( 1 \leq i \leq n \), and the elements of the sets are \( n^2 \) consecutive integers.

8. **Example for \( n = 4 \):**
   \[
   \begin{array}{|c|c|c|c|c|}
   \hline
   & B_1 & B_2 & B_3 & B_4 \\
   \hline
   A_1 & 1 & 2 & 6 & 11 \\
   \hline
   A_2 & 3 & 8 & 5 & 7 \\
   \hline
   A_3 & 4 & 13 & 9 & 16 \\
   \hline
   A_4 & 15 & 12 & 10 & 14 \\
   \hline
   \end{array}
   \]

9. **General case:**
   For \( n \times n \) grid, the minimal difference can be achieved by:
   \[
   A_i = \{(i-1)n + x \mid 1 \leq x \leq n\}
   \]
   \[
   B_i = \{i + xn \mid 0 \leq x \leq n-1\}
   \]

10. **Conclusion:**
    The least possible value of the greatest of all such differences is \( n \).

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