Let $n>0$ and $X_1,X_2,\ldots,X_{n+1}$ are nonempty subsets of $\{1,2,\ldots,n\}$. Prove that there exists two nonempty disjoint subsets $I$ and $J$ of $\{1,2,\ldots,n+1\}$ such that $$\bigcup_{i\in I} X_i=\bigcup_{j\in J} X_j$$ China Western Mathematical Olympiad 2002
Say $v_i$ is indicator vector for the set $X_i$. So, if $X_1= \{1,4,5\}$ then $v_1= (1,0,0,1,1,0,0,...,0,0)$.
Clearly $v_1,...v_{n+1}$ are not independant in $\mathbb{Q}^n$. So there are scalars, not all $0$, such that $$a_1v_1+a_2v_2+...+a_{n+1}v_{n+1}=0$$ Let $P$ be a set of indices of positive scalars and $N$ of negative one, so $P\cap N = \emptyset$. Then we have $$\sum_{i\in P} a_iv_i = -\sum_{j\in N} a_jv_j$$
Now let $\mathcal{P} := \bigcup_{i\in P} X_i$ and $\mathcal{N} := \bigcup_{j\in N} X_j$. It is enough to prove $\mathcal{P}= \mathcal{N}$ and we are done.
If $k\in \mathcal{P}$, then $k$-th coordinate of at least one $v_i$, $i\in P $ is not $0$, but then $k$-th coordinate of sum $-\sum_{j\in N} a_jv_j$ is also not zero, so there must be a vector $v_j$, $j\in N $ with nonzero $k$-th coordinate, so $k\in \mathcal{N}$ and thus $\mathcal{P}\subseteq \mathcal{N}$, since $k$ was arbitrary. Since the other inclusion is now also obivous the claim is proven.