Let $n\geq 2$ be a fixed positive integer. Let $\{a_1,a_2,...,a_n\}$ be fixed positive integers whose sum is $2n-1$. Denote by $S_{\mathbb{A}}$ the sum of elements of a set $A$. Find the minimal and maximal value of $S_{\mathbb{X}}\cdot S_{\mathbb{Y}}$ where $\mathbb{X}$ and $\mathbb{Y}$ are two sets with the property that $\mathbb{X}\cup \mathbb{Y}=\{a_1,a_2,...,a_n\}$ and $\mathbb{X}\cap \mathbb{Y}=\emptyset.$

[i]Note: $\mathbb{X}$ and $\mathbb{Y}$ can have multiple equal elements. For example, when $n=5$ and $a_1=...=a_4=1$ and $a_5=5$, we can consider $\mathbb{X}=\{1,1,1\}$ and $\mathbb{Y}=\{1,5\}$. Moreover, in this case, $S_\mathbb{X}=3$ and $S_{\mathbb{Y}}=6.$[/i]
1. **Define the problem and notation:**
   Let \( n \geq 2 \) be a fixed positive integer. Let \( \{a_1, a_2, \ldots, a_n\} \) be fixed positive integers whose sum is \( 2n - 1 \). Denote by \( S_{\mathbb{A}} \) the sum of elements of a set \( \mathbb{A} \). We need to find the minimal and maximal value of \( S_{\mathbb{X}} \cdot S_{\mathbb{Y}} \) where \( \mathbb{X} \) and \( \mathbb{Y} \) are two sets with the property that \( \mathbb{X} \cup \mathbb{Y} = \{a_1, a_2, \ldots, a_n\} \) and \( \mathbb{X} \cap \mathbb{Y} = \emptyset \).

2. **Sum of elements:**
   Given that the sum of all elements is \( 2n - 1 \), we have:
   \[
   S_{\mathbb{X}} + S_{\mathbb{Y}} = 2n - 1
   \]

3. **Maximizing \( S_{\mathbb{X}} \cdot S_{\mathbb{Y}} \):**
   To maximize the product \( S_{\mathbb{X}} \cdot S_{\mathbb{Y}} \), we need to balance the sums \( S_{\mathbb{X}} \) and \( S_{\mathbb{Y}} \) as much as possible. The optimal balance occurs when one of the sets has a sum of \( n \) and the other has a sum of \( n-1 \). This is because:
   \[
   S_{\mathbb{X}} \cdot S_{\mathbb{Y}} = n \cdot (2n - 1 - n) = n \cdot (n - 1) = n(n-1)
   \]

4. **Existence of such a subset:**
   We need to show that there exists a subset \( T \subset \{a_1, a_2, \ldots, a_n\} \) such that \( \sum_{a_i \in T} a_i = n \). Consider the partial sums \( s_k = a_1 + a_2 + \cdots + a_k \) for \( 1 \leq k \leq n \). If two of these partial sums have the same residue modulo \( n \), then the difference between these sums is a multiple of \( n \). Since the total sum is \( 2n - 1 \), which is less than \( 2n \), the difference must be exactly \( n \). If all partial sums are different modulo \( n \), then one of them must be congruent to \( 0 \pmod{n} \), implying that the sum is exactly \( n \).

5. **Minimizing \( S_{\mathbb{X}} \cdot S_{\mathbb{Y}} \):**
   To minimize the product \( S_{\mathbb{X}} \cdot S_{\mathbb{Y}} \), we should make one of the sums as small as possible. Assume \( a_1 \leq a_2 \leq \cdots \leq a_n \). Clearly, \( a_1 = 1 \) since the sum of all elements is \( 2n - 1 \). Choose \( \mathbb{X} = \{a_1\} \) and \( \mathbb{Y} = \{a_2, a_3, \ldots, a_n\} \). Then:
   \[
   S_{\mathbb{X}} = 1 \quad \text{and} \quad S_{\mathbb{Y}} = 2n - 2
   \]
   Thus:
   \[
   S_{\mathbb{X}} \cdot S_{\mathbb{Y}} = 1 \cdot (2n - 2) = 2n - 2
   \]

Conclusion:
The minimal value of \( S_{\mathbb{X}} \cdot S_{\mathbb{Y}} \) is \( 2n - 2 \) and the maximal value is \( n(n-1) \).

The final answer is \( \boxed{2n - 2 \text{ and } n(n-1)} \)