8 Given a positive integer $n(\geqslant 2)$, find the minimum value of $|X|$, such that for any $n$ two-element subsets $B_{1}, B_{2}, \cdots, B_{n}$ of the set $X$, there exists a subset $Y$ of $X$ satisfying:
(1) $|Y|=n$;
(2) For $i=1,2, \cdots, n$, we have $\left|Y \cap B_{i}\right| \leqslant 1$.
Here $|A|$ denotes the number of elements in the finite set $A$. (Leng Gangsong)
Solve $|X|_{\min }=2 n-1$.
(1) When $|X|=2 n-2$, it is not necessarily true that there exists a $Y$ that satisfies the conditions. In fact, let $X=\{1,2, \cdots, 2 n-2\}$, and consider a partition of $X$ as $B_{1}=\{1,2\}, B_{2}=\{3,4\}, \cdots, B_{n-1}=\{2 n-3,2 n-2\}$. Since $|Y|=n$, there must be at least two elements in $Y$ that belong to the same $B_{j}$, hence $\left|Y \cap B_{j}\right|>1$, which is a contradiction.
(2) We now prove that $|X|=2 n-1$ satisfies the condition.
Let $B=\bigcup_{i=1}^{n} B_{i},|B|=2 n-1-z$, then there are $z$ elements that do not appear in any $B_{i}$, denoted as $a_{1}, a_{2}, \cdots, a_{z}$. If $z \geqslant n-1$, then taking $Y=\left\{a_{1}, \cdots, a_{n-1}, d\right\}, d \in B$ will suffice.
Assume $z<n-1$.
Suppose there are $t$ elements that appear exactly once in $B_{1}, B_{2}, \cdots, B_{n}$. Since $\sum_{i=1}^{n}\left|B_{i}\right|=2 n$, we have
$$
t+2(2 n-1-z-t) \leqslant 2 n,
$$

Thus,
$$
t \geqslant 2 n-2-2 z.
$$

Therefore, the elements that appear at least twice in $B_{1}, B_{2}, \cdots, B_{n}$ appear at most $2 n-(2 n-2-2 z)=2+2 z$ times in total.

Consider the elements $b_{1}, b_{2}, \cdots, b_{t}$ that appear exactly once in $B_{1}, B_{2}, \cdots, B_{n}$. The number of $B_{j}$ that do not contain $b_{1}, b_{2}, \cdots, b_{t}$ is at most $\frac{2+2 z}{2}=1+z$. Hence, at least $n-(z+1)=n-z-1$ of the $B_{j}$ contain $b_{1}, b_{2}, \cdots, b_{t}$.

Without loss of generality, assume $B_{1}, B_{2}, \cdots, B_{n-1-z}$ each contain one of the elements $\tilde{b}_{1}, \widetilde{b}_{2}, \cdots, \tilde{b}_{n-1-z}$ from $b_{1}, b_{2}, \cdots, b_{t}$ (if a $B_{l}(1 \leqslant l \leqslant n-1-z)$ contains multiple, choose one).

Since $2(n-1-z)+z=2 n-2-z<2 n-1$, there must be an element $d$ that does not appear in $B_{1}, B_{2}, \cdots, B_{n-1-z}$ but appears in $B_{n-z}, \cdots, B_{n}$. Let $Y=\left\{a_{1}, a_{2}, \cdots, a_{z}, \widetilde{b}_{1}, \widetilde{b}_{2}, \cdots, \widetilde{b}_{n-1-z}, d\right\}$, then $Y$ satisfies the requirements.