The 37th Korean Mathematical Olympiad Final Round

The answer is $n$. If Alice puts one candy in each of boxes $B_1,B_2,\dots ,B_{4n-1}$, then Bob can choose at most $n$ out of the $2n$ boxes, so Alice can get exactly $n$ candies.

Now we prove that Bob can take at least $n$ candies. Let $b_i$ be the number of candies Alice puts in $B_i$ for $i=1,2,\dots ,4n$.

Suppose there exists $m=1,2,\dots ,n$ such that $b_{4m-2}<2$. We consider the following two sequences $$p_i=\{\begin{array}{ll}1& i=1\\ 1& i=2,4,\dots ,2m-2\\ 3& i=3,5,\dots ,2m-1\\ 3& i=2m,2m+2,\dots ,2n\\ 1& i=2m+1,2m+3,\dots ,2n-1\end{array},q_i=\{\begin{array}{ll}3& i=1,3,\dots ,2m-1\\ 1& i=2,4,\dots ,2m-2\\ 3& i=2m,2m+2,\dots ,2n-2\\ 1& i=2m+1,2m+3,\dots ,2n-1\\ 1& i=2n\end{array}$$ and let $P_i=p_1+p_2+\cdots +p_i$ and $Q_i=q_1+q_2+\cdots +q_i$ for $i=1,2,\dots ,2n$. Then, $\{P_i{\}}_{i=1,2,\dots ,2n}$ and $\{Q_i{\}}_{i=1,2,\dots ,2n}$ satisfy the condition in the problem, and furthermore $\{P_1,P_2,\dots ,P_{2n}\}\cup \{Q_1,Q_2,\dots ,Q_{2n}\}=\{1,2,\dots ,4n\}-\{4m-2\}$. So, $$\sum _{i=1}^{2n}(b_{P_i}+b_{Q_i})\ge 2n-b_{4m-2}\ge 2n-1$$ which implies that either ${\sum }_{i=1}^{2n}{b}_{P_i}$ or ${\sum }_{i=1}^{2n}{b}_{Q_i}$ is at least $n$, so Bob can take at least $n$.

Now we assume that $b_{4m-2}=2$ for every $m=1,2,\dots ,n$. Then, Bob chooses $B_{4k-3},B_{4k-2}$ for $k=1,2,\dots ,n-1$ and $B_{4n-3},B_{4n}$ where Bob takes $2n-2(\ge n)$.