Four, (50 points) Given several rectangular boxes, the lengths of their edges are positive integers not greater than an odd positive integer $n$ (allowing the three edge lengths to be the same), and the thickness of the box walls is negligible. Each box has its three pairs of opposite faces painted red, blue, and yellow, respectively. If no box can fit into another box with the same colored faces parallel, then these boxes are called "harmonious." Find the maximum number of harmonious boxes.
Let the maximum number of harmonious boxes be $f(n)$.

In a three-dimensional Cartesian coordinate system, the coordinate planes
$$
\begin{array}{l}
\{(x, y, z) \mid x=0, y \geqslant 0, z \geqslant 0\}, \\
\{(x, y, z) \mid x \geqslant 0, y=0, z \geqslant 0\}, \\
\{(x, y, z) \mid x \geqslant 0, y \geqslant 0, z=0\}
\end{array}
$$

are colored red, blue, and yellow, respectively. Placing the boxes into the coordinate system such that the three faces sharing a common vertex are placed on the same-colored coordinate planes, each box corresponds to a unique ordered triplet of positive integers $(a, b, c)$, where $a, b, c \in \mathbf{N}_{+}$ and $a, b, c \leqslant n$.

Thus, the set of harmonious boxes corresponds to a point set $D$ in the three-dimensional Cartesian coordinate system,
$$
D \subset\left\{(a, b, c) \mid a, b, c \leqslant n, a, b, c \in \mathbf{N}_{+}\right\},
$$

satisfying that for any two points $\left(a_{1}, b_{1}, c_{1}\right)$ and $\left(a_{2}, b_{2}, c_{2}\right)$ in $D$, we have
$$
\left(a_{1}-a_{2}, b_{1}-b_{2}, c_{1}-c_{2}\right) \notin E,
$$

where $E=\{(x, y, z),(-x,-y,-z) \mid x, y, z \in \mathbf{N}\}$.
$$
\begin{array}{l}
\text { Let } D=\bigcup_{i=1}^{n} D_{i}, \text { where, } \\
D_{i}=\{(a, b, c) \in D \mid c=i\}(i=1,2, \cdots, n) . \\
\text { Let } F_{i}=\left\{(a, b) \mid(a, b, c) \in D_{i}\right\}, \\
f_{i}=\left|F_{i}\right|(i=1,2, \cdots, n) .
\end{array}
$$

Clearly, any two points in $D$ have at most one coordinate in common, i.e.,
$$
F_{i} \cap F_{j}=\varnothing(1 \leqslant i < j \leqslant n) .
$$

For any $F_{i}$, we can assume that the points in $F_{i}$ are ordered such that
$$
\begin{array}{l}
a_{1}>a_{2}>\cdots>a_{f_{i}} \geqslant 1, \\
b_{1}>b_{2}>\cdots>b_{f_{i}} \geqslant 1 .
\end{array}
$$

Thus, $f_{i} \leqslant \min \left\{b_{1}, n+1-a_{1}\right\}$.
In the $(x, y)$ coordinate plane, let
$$
\begin{array}{l}
G_{i}=\int_{j=1}^{f_{i}}\left\{(x, y) \mid x \leqslant a_{j}, y \leqslant b_{j}, x, y \in \mathbf{N}_{+}\right\}, \\
\left.H_{i}=\bigcup_{j=0}^{f_{i}}\left\{(x, y) \mid a_{j} < x \leqslant a_{j+1}, b_{j+1} < y \leqslant b_{j}, x, y \in \mathbf{N}_{+}\right\}\right\},
\end{array}
$$

where $a_{0}=0, b_{0}=n+1, a_{f_{i}+1}=n+1, b_{f_{i}+1}=0$.
Thus, equation (2) is equivalent to
$$
G_{i} \cap F_{k}=\varnothing(k=i-1, i-2, \cdots, 1) .
$$

For $i=n, n-1, \cdots, 2$, if $F_{i-1} \cap H_{i} \neq \varnothing$, then merge all points in $F_{i-1} \cap H_{i}$ into $F_{i}$. Denote the new set as $F_{i}^{\prime}$, then $\left\{F_{i}^{\prime}\right\}$ still satisfies equation (3).
Continue the adjustment for $\left\{F_{i}^{\prime}\right\}$ until
$$
F_{i-1}^{\prime} \cap H_{i}^{\prime}=\varnothing(i=n, n-1, \cdots, 2) \text {. }
$$

At this point, still denote it as $\left\{F_{i}\right\}$.
Connect the points in each $F_{i}$ to form a broken line, then all $F_{i}$
form $n$ non-intersecting broken lines in
$$
\{(x, y) \mid 1 \leqslant x \leqslant n, 1 \leqslant y \leqslant n\}
$$

Thus,
$$
\begin{array}{l}
\text { we have } f(n)=|D|=\sum_{i=1}^{n}\left|D_{i}\right|=\sum_{i=1}^{n}\left|F_{i}\right| \\
\leqslant n+2\left[(n-1)+(n-2)+\cdots+\frac{n+1}{2}\right] \\
=n+\frac{n-1}{2} \cdot \frac{3 n-1}{2}=\frac{3 n^{2}+1}{4} . \\
\text { When } F_{i}=\left\{(a, b) \left\lvert\, a+b=\frac{3(n+1)}{2}-i\right.\right\}(i=1,2, \cdots, n),
\end{array}
$$

i.e., $D=\left\{(a, b, c) \left\lvert\, a+b+c=\frac{3(n+1)}{2}\right., a, b, c \in \mathbf{N}_{+}\right\}$,
the equality holds.
In summary, the maximum number of harmonious boxes is $\frac{3 n^{2}+1}{4}$.