Brazilian Math Olympiad

Since $2011=33\cdot 61-2$, consider the example where $S_i=\{k\in \mathbb{Z}\mid 61(i-1)<k\le 61i\}$ for $i=1,2,\dots ,31$, $S_{32}=\{k\in \mathbb{Z}\mid 61\cdot 31<k<61\cdot 32\}$ and $S_{33}=\{k\in \mathbb{Z}\mid 61\cdot 32\le k\le 2011\}$. Notice that $|S_i|=61$ for $1\le i\le 31$ and $|S_i|=60$ for $i=32$ and $i=33$. Thus $|S_i\cup S_j|\le 2\cdot 61$, and therefore $m>2011-2\cdot 61=1889$.

Now we prove that the minimum value of $m$ is, in fact, $m=1890$. First, notice that if $m=1890$ then $|S_i\cup S_j|\le 2011-1890=121$. Suppose that $|S_1\cup S_2\cup S_3|>181$. Then one of the sets, say $S_1$, has more than $181/3$ elements, that is, $|S_1|\ge 61$. But $|(S_1\cup S_2\cup S_3)\setminus (S_1\cup S_2)|>181-121⟺|S_3\setminus (S_1\cup S_2)|>60$. But $|S_3\cup S_1|=|S_3\setminus S_1|+|S_1|\ge |S_3\setminus (S_1\cup S_2)|+|S_1|>60+61=121$, contradiction. Hence $|S_1\cup S_2\cup S_3|\le 181$. So, $|S_1\cup S_2\cup \cdots \cup S_{33}|\le |S_1\cup S_2\cup S_3|+|S_4\cup S_5|+|S_6\cup S_7|+\cdots +|S_{32}+S_{33}|\le 181+15\cdot 121=1996$, and there exists sixteen stickers that none of the 33 friends have.

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Alternative solution.

We will prove that $m=1890$ in another way. The example for $m=1889$ is the same from the previous solution. Again, number the stickers from 1 to 2011 and let $T_i$ be the set of the stickers that the friend $i$ does not have, $1\le i\le 33$. Consider all pairs $(x,\{i,j\})$ such that $x\in T_i\cap T_j$. If for every sticker there is a friend that has it, that is, $T_1\cap T_2\cap \cdots \cap T_{33}=\varnothing$ then for each $x$ there exists $k$ such that $x\notin T_k$. So each $x$ belongs to at most 32 sets $T_i$ and, hence, there exist at most $2011\cdot \left(\genfrac{}{}{0ex}{}{32}{2}\right)$ pairs $(x,\{i,j\})$. On the other hand, since $|T_i\cap T_j|\ge m$ there exists at least $m\cdot \left(\genfrac{}{}{0ex}{}{33}{2}\right)$ pairs $(x,\{i,j\})$. Therefore, if $T_1\cap T_2\cap \cdots \cap T_{33}=\varnothing$ then $$m\cdot \left(\genfrac{}{}{0ex}{}{33}{2}\right)\le 2011\cdot \left(\genfrac{}{}{0ex}{}{32}{2}\right)⟺m\le 1890$$