The 62nd question: Given a positive integer $n \geq 2$, find the minimum value of $|X|$ such that for any $n$ two-element subsets $B_{1}$, $B_{2}$, ..., $B_{n}$ of the set $X$, there exists a subset $Y$ of $X$ satisfying:
(1) $|Y|=n$;
(2) For $i=1$, $2$, ..., $n$, $\left|Y \cap B_{i}\right| \leq 1$.
The 62nd question:
Solution Method One: The minimum value of $|X|$ is $2n-1$.
(1) First, prove that when $|X|=2n-2$, there may not exist a set $Y$ that satisfies the conditions.

In fact, let $X=\{1,2, \ldots, 2n-2\}$. Consider a partition of $X$: $B_{i}=\{2i-1,2i\} \quad 1 \leq i \leq n-1$. Since $|Y|=n$, by the pigeonhole principle, there must be two elements in $Y$ that belong to the same $B_{j}$, hence $|Y \cap B_{j}|=2$, which contradicts the condition.
(2) Next, prove that when $|X|=2n-1$, there must exist a set $Y$ that satisfies the conditions.
Let $B=\cup_{i=1}^{n} B_{i}, |B|=2n-1-z$, then there are $z$ elements that do not appear in any $B_{i}$, denoted as $a_{1}, a_{2}, \ldots, a_{z}$. If $|z| \geq n-1$, then take $Y=\{a_{1}, a_{2}, \ldots, a_{n-1}, d\} (d \in B)$.
Assume $z < n-1$.
Let the number of elements that appear exactly once in $B_{1}, B_{2}, \ldots, B_{n}$ be $t$. Since $\sum_{i=1}^{n}|B_{i}|=2n$, we have $t + 2(2n-1-z-t) \leq 2n$
$$
\Rightarrow t \geq 2n-2-2z
$$

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

Consider the elements $b_{1}, b_{2}, \ldots, b_{t}$ that appear exactly once in $B_{1}, B_{2}, \ldots, B_{n}$. The number of $B_{j}$ that do not contain $b_{1}, b_{2}, \ldots, b_{t}$ is at most $\frac{2+2z}{2}=1+z$. Therefore, at least $n-(z+1)=n-z-1$ of the $B_{j}$ contain $b_{1}, b_{2}, \ldots, b_{t}$. Without loss of generality, assume $B_{1}, B_{2}, \ldots, B_{n-1-z}$ each contain one of $b_{1}, b_{2}, \ldots, b_{n-1-2}$ (if there are multiple such $B_{j}(1 \leq j \leq n-z-1)$, choose one).

Since $2(n-1-z)+z=2n-2-z<2n-1$, there must be an element $d$ that does not appear in $B_{1}, B_{2}, \ldots, B_{n-1-2}$ but appears in $B_{n-2}, B_{n-z+1}, \ldots, B_{n}$. Take
$$
Y=\{a_{1}, a_{2}, \ldots, a_{z}, b_{1}, b_{2}, \ldots, b_{n-z-1}, d\}
$$

which satisfies the conditions.
In summary, the minimum value of $|X|$ is $2n-1$.
Solution Method Two: The minimum value of $|X|$ is $2n-1$.
(1) First, prove that when $|X|=2n-2$, there may not exist a set $Y$ that satisfies the conditions.
In fact, let $X=\{1,2, \ldots, 2n-2\}$. Consider a partition of $X$: $B_{i}=\{2i-1,2i\} (1 \leq i \leq n-1)$. Since $|Y|=n$, by the pigeonhole principle, there must be two elements in $Y$ that belong to the same $B_{j}$, hence $|Y \cap B_{j}|=2$, which contradicts the condition.
(2) Next, prove by induction on $n$ that when $|X|=2n-1$, there must exist a set $Y$ that satisfies the conditions.
When $n=2$, the conclusion is obviously true.
Assume the conclusion holds for $n-1 (n \geq 3)$. Consider the case for $n$.
Construct a bipartite graph $G(U, V)$, where $U$ has $2n-1$ points representing the $2n-1$ elements of $X$, and $V$ has $n$ points representing $B_{1}, B_{2}, \ldots, B_{n}$. If an element in $X$ belongs to a $B_{i}$, connect the corresponding points with an edge; otherwise, do not connect them. According to the conditions, for any $v \in V$, $d(v)=2$.
Let the number of points in $U$ with degree 0 be $x$, the number of points with degree 1 be $y$, and the number of points with degree at least 2 be $z$, then
$$
\begin{array}{l}
\left\{\begin{array}{c}
x+y+z=2n-1 \\
0 \cdot x+1 \cdot y+2 \cdot z \leq \sum_{u \in U} d(u)=\sum_{v \in V} d(v)=2n
\end{array}\right. \\
\Rightarrow 2x+y \geq 2n-2 \\
\end{array}
$$
(1) If $x \geq n-1$, let the $n-1$ points with degree 0 correspond to elements $a_{1}, a_{2}, \ldots, a_{n-1}$, then take $Y=\{a_{1}, a_{2}, \ldots, a_{n-1}, d\} (d \in B)$, which satisfies the conditions.
(2) If $x \leq n-2$, then $y \geq 2$. Choose a point with degree 1, let the corresponding element be $a$, and assume $a$ belongs only to the subset $B_{n}$, and let $B_{n}=\{a, b\}$.

Let $X' = X \setminus \{a, b\}$, and let $B_{i}' = B_{i} \setminus \{b\} (1 \leq i \leq n-1)$. Then $|X'|=2(n-1)-1$, and each $B_{i}'$ is a subset of $X'$, with $|B_{i}'| \leq 2 (1 \leq i \leq n-1)$. According to the induction hypothesis, we can select $n-1$ elements from $X'$ to form a set $Y'$ such that $|Y' \cap B_{i}'| \leq 1$ for any $1 \leq i \leq n-1$.
Then take $Y = Y' \cup \{a\}$, so $|Y|=n$, and for each $1 \leq i \leq n$, $|Y \cap B_{i}| \leq 1$.
In summary, the minimum value of $|X|$ is $2n-1$.