4. (10 points) In a deck of cards, $J, Q, K$ are considered as $11,12,13$ points respectively. From 13 cards numbered from 1 to 13, what is the maximum number of cards you can pick so that there are no 2 pairs of cards, where the sum of the points of one pair equals the sum of the points of the other pair?
4. (10 points) In a deck of cards, $J, Q, K$ are considered as 11, 12, 13 points respectively. From 13 cards numbered from 1 to 13, what is the maximum number of cards that can be selected so that there are no 2 pairs of cards, where the sum of the points of one pair equals the sum of the points of the other pair? 【Analysis】First, we get that in the 13 cards numbered from 1 to 13, there are differences from $1 \sim 12$. Further, we find that there are at least 5 groups where $2 a_{n}=a_{m}+a_{l}$ $(m>n>l)$. This is equivalent to selecting these 5 groups from 21 groups, leaving 16 groups. Then, using the pigeonhole principle, we can solve the problem.
【Solution】Let's assume at least $n$ cards are selected, but in the set $\{1,2,3,5,8,13\}$, there are no 2 pairs of cards where the sum of the points of one pair equals the sum of the points of the other pair, so $n \geqslant 6$. If $n=7$, the 7 selected cards, arranged from smallest to largest, are: $\left\{a_{1}, a_{2}, \cdots, a_{7}\right\}$. For any 2 cards $a_{m}$ and $a_{n}$, where $a_{m}>a_{n}$,
then $\frac{7 \times 6}{2}=21$ differences in the range $1 \sim 12$, and these differences:
(1) cannot have $a_{m}-a_{n}=a_{k}-a_{n}$,
(2) if $a_{m}-a_{n}=a_{n}-a_{l}$, i.e., $2 a_{n}=a_{m}+a_{l}(m>n>l)$,
then there cannot be $m_{1} \neq m, l_{1} \neq l\left(m_{1}>n>l_{1}\right)$ such that $2 a_{n}=a_{m 1}+a_{l 1}$, otherwise there would be 2 pairs of cards where the sum of the points of one pair equals the sum of the points of the other pair.
That is, for the same $n$, the occurrence of $a_{m}-a_{n}=a_{n}-a_{l}$ or $2 a_{n}=a_{m}+a_{l}(m>n>l)$ is unique. There are at least 5 groups where $2 a_{n}=a_{m}+a_{l}(m>n>l)$, which is equivalent to selecting these 5 groups from 21 groups, leaving 16 groups. The 16 values in the range $1 \sim 12$ must have at least two that are the same.