6. There are several children, the first child has 2 more candies than the second child, the second child has 2 more candies than the third child, “, that is, each preceding child always has 2 more candies than the next child. They sit in a circle in order to play a game, starting with the first child giving 2 candies to the second child, the second child giving 4 candies to the third child, $\cdots$, that is, each child always passes on the candies received from the previous child plus 2 more of their own candies to the next child. When the game reaches a point where a child receives candies from the previous child but cannot pass on the required number of candies, the ratio of candies between two adjacent children is $13 ： 1$. How many children are there at most?
6. 【Solution】Suppose there are $\mathrm{n}$ children, passing the candy for a total of $\mathrm{k}$ rounds (the last child passes $\mathrm{k}-1$ rounds), and the number of candies each person has when the game is interrupted is $\mathrm{a}$. First, we study the value of $\mathrm{a}$. When the game is interrupted at 0 (the last child has no candy to pass), $a_{n}=2 \mathrm{nk}-2, a_{n-1}=0, a_{1}=2 \mathrm{n}-4$; when the game is interrupted at 1 (the last child has only one candy), $a_{n}=2 \mathrm{nk}-1, a_{n-1}=1, a_{1}=2 \mathrm{n}-3$. We discuss six cases:
(1) 0 interruption, $a_{n}: a_{n-1}=13: 1$, i.e., $\frac{2 n k-2}{0}=\frac{13}{1}$, which clearly has no solution.
(2) 0 interruption, $a_{n}: a_{1}=13: 1$, i.e., $\frac{2 n k-2}{2 n-4}=\frac{13}{1} \Rightarrow 26 n-52=2 n k-2 \Rightarrow n (13-\mathrm{k})=25$,

which gives $n=25, k=12 (n=5, k=8$ is discarded)
(3) 0 interruption, $a_{1}: a_{n}=13: 1$, i.e., $\frac{2 n-4}{2 n k-2}=\frac{13}{1} \Rightarrow 26 n k-26=2 n-4 \Rightarrow n (13 \mathrm{k}-1)=11$, which has no integer solution.
(4) 1 interruption, $a_{n}: a_{n-1}=13: 1$, i.e., $\frac{2 n k-1}{1}=\frac{13}{1} \Rightarrow 2 n k-1=13 \Rightarrow n k =7$,

which gives $n=7, k=1 (n=1, k=7$ is discarded)
(5) 1 interruption, $a_{n}: a_{1}=13: 1$, i.e., $\frac{2 n k-1}{2 n-3}=\frac{13}{1} \Rightarrow 26 n-39=2 n k-1 \Rightarrow n (13-\mathrm{k})=19$,

which gives $n=19, k=12$
(6) 1 interruption, $a_{1}: a_{n}=13: 1$, i.e., $\frac{2 n-3}{2 n k-1}=\frac{13}{1} \Rightarrow 26 n k-13=2 n-3 \Rightarrow n (13 \mathrm{k}-1)=5$, which has no integer solution.

From the above analysis, we can conclude that the maximum number of children is 25.