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 $|(S_1 \cup S_2 \cup S_3) \setminus (S_1 \cup S_2)| > 181 - 121 \iff |S_3 \setminus (S_1 \cup S_2)| > 60$. 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, $|S_1 \cup S_2 \cup \dots \cup S_{33}| \le |S_1 \cup S_2 \cup S_3| + |S_4 \cup S_5| + |S_6 \cup S_7| + \dots + |S_{32} + S_{33}| \le 181 + 15 \cdot 121 = 1996$, and there exists sixteen stickers that none of the 33 friends have.

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, $T_1 \cap T_2 \cap \dots \cap T_{33} = \emptyset$ 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 $2011 \cdot \binom{32}{2}$ pairs $(x, \{i, j\})$. On the other hand, since $|T_i \cap T_j| \ge m$ there exists at least $m \cdot \binom{33}{2}$ pairs $(x, \{i, j\})$. Therefore, if $T_1 \cap T_2 \cap \dots \cap T_{33} = \emptyset$ then
$$m \cdot \binom{33}{2} \le 2011 \cdot \binom{32}{2} \iff m \le 1890$$