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
$$p_i = \begin{cases} 1 & i=1 \\ 1 & i=2, 4, \dots, 2m-2 \\ 3 & i=3, 5, \dots, 2m-1 \\ 3 & i=2m, 2m+2, \dots, 2n \\ 1 & i=2m+1, 2m+3, \dots, 2n-1 \end{cases}, \quad q_i = \begin{cases} 3 & i=1, 3, \dots, 2m-1 \\ 1 & i=2, 4, \dots, 2m-2 \\ 3 & i=2m, 2m+2, \dots, 2n-2 \\ 1 & i=2m+1, 2m+3, \dots, 2n-1 \\ 1 & i=2n \end{cases}$$
and let $P_i = p_1 + p_2 + \dots + p_i$ and $Q_i = q_1 + q_2 + \dots + q_i$ 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 $2n - 2 (\ge n)$.