Given a positive integer $ n\geq 2$, let $ B_{1}$, $ B_{2}$, ..., $ B_{n}$ denote $ n$ subsets of a set $ X$ such that each $ B_{i}$ contains exactly two elements. Find the minimum value of $ \left|X\right|$ such that for any such choice of subsets $ B_{1}$, $ B_{2}$, ..., $ B_{n}$, there exists a subset $ Y$ of $ X$ such that:
 (1) $ \left|Y\right| \equal{} n$;
 (2) $ \left|Y \cap B_{i}\right|\leq 1$ for every $ i\in\left\{1,2,...,n\right\}$.
1. **Initial Setup and Problem Restatement**:
   We are given a set \( X \) and \( n \) subsets \( B_1, B_2, \ldots, B_n \) of \( X \), each containing exactly two elements. We need to find the minimum value of \( |X| \) such that there exists a subset \( Y \subseteq X \) with \( |Y| = n \) and \( |Y \cap B_i| \leq 1 \) for all \( i \in \{1, 2, \ldots, n\} \).

2. **Lower Bound Argument**:
   Consider \( |X| = 2n - 2 \). We construct the subsets \( B_i \) as follows:
   \[
   B_1 = \{1, 2\}, B_2 = \{3, 4\}, \ldots, B_{n-1} = \{2n-3, 2n-2\}, B_n = \{1, 3\}
   \]
   In this case, any subset \( Y \) of \( X \) with \( |Y| = n \) will have at least one \( B_i \) such that \( |Y \cap B_i| \geq 2 \). Therefore, \( |X| = 2n - 2 \) is not sufficient.

3. **Proving \( |X| \geq 2n - 1 \)**:
   We need to show that \( |X| \geq 2n - 1 \) is necessary. Assume \( |X| = 2n - 1 \). We will prove that there exists a subset \( Y \subseteq X \) with \( |Y| = n \) and \( |Y \cap B_i| \leq 1 \) for all \( i \).

4. **Constructing the Subset \( Y \)**:
   Choose \( Y \) such that \( |Y| \) is maximized and \( |Y \cap B_i| \leq 1 \) for all \( i \). Suppose \( |Y| \leq n - 1 \). Then \( |X \setminus Y| \geq n \).

5. **Contradiction Argument**:
   For each \( a \in X \setminus Y \), there exists \( 1 \leq t(a) \leq n \) and \( x_a \in Y \) such that \( (a, x_a) \in B_{t(a)} \). If not, we could add \( a \) to \( Y \) and get \( |Y'| > |Y| \), contradicting the maximality of \( |Y| \).

6. **Distinct Subsets**:
   If \( a \neq b \), then \( B_{t(a)} \neq B_{t(b)} \). Since \( |X \setminus Y| \geq n \), we have at least \( n \) distinct subsets \( B_{t(a)} \). Thus, \( |X \setminus Y| = n \).

7. **Final Contradiction**:
   The subset \( X \setminus Y \) satisfies \( |(X \setminus Y) \cap B_i| = 1 \) for all \( i \) and \( |X \setminus Y| = n \), which is a contradiction. Therefore, \( |Y| \) must be exactly \( n \).

8. **Conclusion**:
   Hence, the minimum value of \( |X| \) such that there exists a subset \( Y \subseteq X \) with \( |Y| = n \) and \( |Y \cap B_i| \leq 1 \) for all \( i \) is \( 2n - 1 \).

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