It is known that the difference between any two of the five numbers $1, 4, 6, 7, 9$ is not 4 or 7. Adding 11 to each of these numbers gives $12, 15, 17, 18, 20$. Clearly, the difference between any two of these five numbers is also not 4 or 7, and the difference between any of these numbers and any of the first five numbers is also not 4 or 7. By analogy, for each fixed positive integer $n$, five numbers can be selected from every consecutive 11 numbers. The remaining $i=n-11 \times\left[\frac{n}{11}\right]$ (where $[x]$ denotes the greatest integer not exceeding the real number $x$) numbers can have $R_{i}$ numbers selected, forming the set $T_{n}$.

We now prove that $T_{n}$ cannot contain more numbers.
If not, then at least one of the 11-number groups in the above $T_{n}$ can have six numbers selected such that the difference between any two of these numbers is not 4 or 7.

Consider the group of numbers $1, 2, \cdots, 11$, and divide them into five subgroups:
$(1,8),(2,6),(3,10),(5,9),(4,7,11)$. At least one of these subgroups must have two numbers selected.

Clearly, the two numbers in each of the first four subgroups cannot both be selected; at most one number can be selected from each subgroup. Therefore, the only possibility is to select $4$ and $7$ from the fifth subgroup $(4,7,11)$. Thus, from $(1,8)$, only $1$ can be selected; from $(5,9)$, only $9$ can be selected; and from $(2,6)$, only $6$ can be selected.

Then, from $(3,10)$, neither $3$ nor $10$ can be selected. Hence, it is impossible to select a sixth number.
Therefore, $f_{n}=5 \times\left[\frac{n}{11}\right]+R_{i}$.
When $i=0,1,2,3$, $R_{i}=i$.
When $i=4,5, \cdots, 8$, for example, when $i=8$, consider the remaining numbers $1,2, \cdots, 8$, and divide them into four groups:
$(1,5),(2,6),(3,7),(4,8)$,

one number can be selected from each group, totaling 4 numbers, i.e., $R_{i}=4$;
When $i=9,10$, for example, when $i=9$, consider the remaining numbers $1,2, \cdots, 9$, and divide them into five groups:
$(1,5),(2,6),(3,7),(4,8),(9)$, one number can be selected from each group, i.e., $1, 3, 4, 6, 9$, so $R_{i}=5$.
Since $2023=11 \times 183+10$, we have
$$
\begin{array}{l}
\sum_{n=1}^{2023} f_{n}=\sum_{k=0}^{183} \sum_{i=0}^{10} f_{11 k+i} \\
=\sum_{k=0}^{183}(11 \times 5 k+1+2+3+5 \times 4+2 \times 5) \\
=55 \times \frac{183 \times 184}{2}+184 \times 36=932604 .
\end{array}
$$
(Liu Xiaojie, Wan Zhaoxun, Shijiazhuang No. 2 High School, Hebei Province, 050004)
Four. (50 points) Let $n$ be a positive integer, and let $T_{n}$ be a subset of the set $A_{n}=\left\{k \mid k \in \mathbf{Z}_{+}\right.$, and $\left.k \leqslant n\right\}$, such that the difference between any two numbers in $T_{n}$ is not equal to 4 or 7. If the maximum number of elements in $T_{n}$ is denoted by $f_{n}\left(\right.$ for example, $\left.f_{1}=1, f_{2}=2\right)$, find $\sum_{n=1}^{2023} f_{n}$.