Four, (50 points) Given a positive integer $n(n \geqslant 3)$. Three types of fruits are distributed into $n$ boxes. Try to find the smallest positive integer $k$, such that no matter how the fruits are distributed, it is always possible to select $k$ boxes, in which the three types of fruits are each at least half of their respective total quantities.
Let's denote three types of fruits as $x, y, z$, and $A_{i}(i=1,2, \cdots, n)$ represents the amount of $x$ fruit in the $i$-th box.

Consider such a distribution: one box contains all the $x$ fruits, another box contains all the $y$ fruits, and the remaining $n-2$ boxes evenly distribute all the $z$ fruits. Clearly,
$$
k \geqslant 1+1+\left[\frac{n-2+1}{2}\right]=\left[\frac{n+3}{2}\right],
$$

where $[x]$ denotes the greatest integer not exceeding the real number $x$.
We will now prove that $\left[\frac{n+3}{2}\right]$ is the minimum value sought.
First, we prove a lemma.
Lemma: In $2 m$ boxes containing $x$ and $y$ fruits, the maximum amount of $x$ fruit in a single box is $x_{0}$, and the maximum amount of $y$ fruit in a single box is $y_{0}$. Then, the boxes can be divided into two groups $S$ and $T$, each containing $m$ boxes, such that
$$
\begin{array}{l}
|S(x)-T(x)| \leqslant x_{0}, \\
|S(y)-T(y)| \leqslant y_{0} .
\end{array}
$$

Proof: Let
$$
x_{0}=A_{1}(x) \geqslant A_{2}(x) \geqslant \cdots \geqslant A_{2 m}(x) \text {. }
$$

Let $S_{1}=\left\{A_{1}\right\}, T_{1}=\left\{A_{2}\right\}$.
Then $\left|S_{1}(y)-T_{1}(y)\right|=\left|A_{1}(y)-A_{2}(y)\right|$
$\leqslant \max \left\{A_{1}(y), A_{2}(y)\right\} \leqslant y_{0}$.
Assume that $A_{1}, A_{2}, \cdots, A_{2 i-1}, A_{2 i}$ have been evenly divided into
$$
\begin{array}{l}
S_{i}=\left\{B_{1}, B_{2}, \cdots, B_{i}\right\}, \\
T_{i}=\left\{C_{1}, C_{2}, \cdots, C_{i}\right\},
\end{array}
$$

satisfying
$$
\left|S_{i}(y)-T_{i}(y)\right| \leqslant y_{0} .
$$

Without loss of generality, assume $S_{i}(y) \geqslant T_{i}(y)$, and the box with fewer $y$ fruits among $A_{2 i+1}$ and $A_{2 i+2}$ is $B_{i+1}$, and the one with more $y$ fruits is $C_{i+1}$.
Let $S_{i+1}=S_{i} \cup\left\{B_{i+1}\right\}$,
$T_{i+1}=T_{i} \cup\left\{C_{i+1}\right\}$.
Then $\left|S_{i+1}(y)-T_{i+1}(y)\right|$
$=\left|S_{i}(y)-T_{i}(y)-\left(C_{i+1}(y)-B_{i+1}(y)\right)\right|$
$$
\leqslant \max \left\{S_{i}(y)-T_{i}(y), C_{i+1}(y)-B_{i+1}(y)\right\}
$$
$\leqslant y_{0}$.
By this method, all $2 m$ boxes are divided into two groups
$$
\begin{array}{l}
S=\left\{B_{1}, B_{2}, \cdots, B_{m}\right\}, \\
T=\left\{C_{1}, C_{2}, \cdots C_{m}\right\},
\end{array}
$$

and $|S(y)-T(y)| \leqslant y_{0}$.
Without loss of generality, assume $S(x) \geqslant T(x)$.
Notice that, by the original sorting,
$$
C_{i}(x) \geqslant B_{i}(x) \text {. }
$$

Then $0 \leqslant S(x)-T(x)$
$$
\begin{array}{l}
=\left(B_{1}(x)-C_{m}(x)\right)-\sum_{i=1}^{m-1}\left(C_{i}(x)-B_{i+1}(x)\right) \\
\leqslant B_{1}(x)-C_{m}(x) \leqslant x_{0} .
\end{array}
$$

Returning to the original problem.
When $n$ is even, assume the first box contains the most $x$ fruits, and among the remaining $n-1$ boxes, the second box contains the most $y$ fruits. The other $n-2$ boxes are divided into two groups $S$ and $T$ according to the lemma.
Assume $S(z) \geqslant T(z)$.
By $S(x)+A_{1}(x) \geqslant T(x)$,
$$
S(y)+A_{2}(y) \geqslant T(y),
$$

we know that $S \cup\left\{A_{1}, A_{2}\right\}$ contains at least half of the total amount of each type of fruit, and the number of boxes is
$$
1+1+\frac{n-2}{2}=\left[\frac{n+3}{2}\right] \text {. }
$$

When $n$ is odd, first select any one box, and the remaining $n-1$ boxes are handled as above, resulting in
$$
1+1+1+\frac{n-3}{2}=\left[\frac{n+3}{2}\right]
$$

boxes, each containing at least half of the total amount of each type of fruit.
Therefore, the minimum value of $k$ is $\left[\frac{n+3}{2}\right]$.
(Li Gang, Nanjing National Laboratory of Microstructures, School of Physics, Nanjing University, 210093)