8. Given a natural number $k$. In the city, there are several children who attend several clubs. It is known that no more than $3 k$ children attend each club, each child attends exactly three clubs, and for any two children, there is a club that both of them attend. What is the maximum number of children that can be in the city? (I. Bogdanov, 5. Chelnokov)
Answer. $7 k$. Solution. Example. Let's divide $7 k$ children into 7 groups (numbered from 1 to 7) with $k$ children in each, and form 7 clubs from children of groups $(1,2,3) ;(1,4,5) ;(1,6,7) ;(2,4,6) ;(2,5,7) ;(3,4,7) ;(3,5,6)$. It is not hard to verify that all conditions are met.

Estimate. If all clubs consist of no more than $3(k-1)$ children, we can replace $k$ with $k-1$ (and prove that the number of children does not exceed $7(k-1)$). Thus, we can assume that there is a club $A$ with at least $3 k-2$ children. We can also assume that there is a child $s$ outside $A$, otherwise the number of children does not exceed $3 k$. Three clubs that $s$ attends cover $A$ (since $s$ has a common club with any child from $A$). Therefore, there is a club different from $A$ and intersecting with $A$ in at least $k$ children.

Let $d$ be the maximum number of children in the intersection of two different clubs; by the above, $d \geq k$. Consider clubs $B$ and $C$ intersecting in $d$ children. Let $X$ be the intersection of clubs $B$ and $C$, and $Y$ be the set of all children who do not attend either $B$ or $C$.

Let $x$ be a child from $X$. He attends $B, C$, and some third club $D_{x}$, which, by the condition, must include all children from $Y$. If for some child $z$ from $X$ his third club $D_{z}$ is different from $D_{x}$, then $D_{z}$ and $D_{x}$ intersect in at least $Y$, so $|Y| \leq d$; thus, the total number of children $|B|+|C|-|X|+|Y|$ does not exceed $3 k+3 k-d+d=6 k$. Otherwise, the club $D_{x}$ contains all children from $X$ and $Y$, so $|X|+|Y| \leq 3 k$, from which the total number of children does not exceed $3 k+3 k-2|X|+(|X|+|Y|) \leq 9 k-2 d \leq 7 k$.

Remark. With the restriction on the club size of $3 k+1$ or $3 k+2$, the answers in the problem would be $7 k+1$ and $7 k+4$, respectively.