China Western Mathematical Olympiad

We first prove that $|X|>2n-2$. In fact, if $|X|=2n-2$, let $X=\{1,2,\dots ,2n-2\}$, $B_1=\{1,2\}$, $B_2=\{3,4\}$, $\dots$, $B_{n-1}=\{2n-3,2n-2\}$. Since $|Y|=n$, there exist two elements in $Y$ that belong to the same $B_i$, then $|Y\cap B_i|>1$, a contradiction.

Let $|X|=2n-1$. Let $B={\bigcup }_{i=1}^n{B}_i$, then $|B|=2n-1-z$, where $z$ is the number of subset $X\setminus B$. Suppose the elements of $X\setminus B$ are $a_1,a_2,\dots ,a_z$.

If $z\ge n-1$, take $Y=\{a_1,\dots ,a_{n-1},d\}$, and $d\in B$, as desired.

If $z<n-1$, suppose there are $t$ elements that occur once in $B_1,B_2,\dots ,B_n$. Since ${\sum }_{i=1}^n|B_i|=2n$, then $$t+2(2n-1-z-t)\le 2n,$$ it follows that $t\ge 2n-2-2z$. So the elements that occur twice or more than twice in $B_1,B_2,\dots ,B_n$ occur repeatedly by $2n-(2n-2-2z)=2+2z$ times.

Consider the elements that occur once in $B_1,B_2,\dots ,B_n$: $b_1,b_2,\dots ,b_t$. Thus, there are at most $\frac{2+2z}{2}=1+z$ subsets in $B_1,B_2,\dots ,B_n$ that do not contain the elements $b_1,b_2,\dots ,b_t$. So, there exist $n-(z+1)=n-z-1$ subsets containing at least the elements $b_1,b_2,\dots ,b_t$.

Suppose that $B_1,B_2,\dots ,B_{n-1-z}$ contain the elements ${\tilde{b}}_1,{\tilde{b}}_2,\dots ,{\tilde{b}}_{n-1-z}$ of $b_1,b_2,\dots ,b_t$, respectively. Since $$2(n-1-z)+z=2n-2-z<2n-1,$$ there must exist an element $d$ that is not in $B_1,B_2,\dots ,B_{n-1-z}$ but is in $B_{n-z},\dots ,B_n$.

Write $Y=\{a_1,\dots ,a_z,{\tilde{b}}_1,{\tilde{b}}_2,\dots ,{\tilde{b}}_{n-1-z},d\}$, as desired.