4. (10 points) In a deck of cards, $J, Q, K$ are considered as 11, 12, 13 points respectively. From the 13 cards numbered 1 to 13, what is the maximum number of cards you can pick such 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, get the 13 cards from 1 to 13 points, which have differences of $1 \sim 12$. Further, we get that the equation $2 a_{n}=a_{m}+a_{l}$ $(m>n>l)$ appears at least 5 times, which is equivalent to selecting these 5 groups from 21 groups, leaving 16 groups. Then, according to the pigeonhole principle, we can solve the problem.

【Solution】Solution: Let at least $n$ cards be picked, 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 cards taken, arranged from smallest to largest, are: $\left\{a_{1}, a_{2}, \cdots, a_{7}\right\}$. For any 2 cards $a_{m}$ and $a_{n}$, let $a_{m}>a_{n}$,

then $\frac{7 \times 6}{2}=21$ differences in the range $1 \sim 12$, among these differences:
(1) It is impossible to 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 exist 2 pairs of cards where the sum of the points of one pair equals the sum of the points of the other pair, i.e., 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. The equation $2 a_{n}=a_{m}+a_{l}(m>n>l)$ appears at least 5 times, which is equivalent to selecting these 5 groups from 21 groups, leaving 16 groups, 16 values in the range $1 \sim 12$, there must be two that are the same.