China-TST-2025A

Let's generalize the problem by replacing $2025$ with $n$ and $66$ with $m$. For any positive integers $m\le n$, define: $$f_m(n)=\underset{\begin{array}{c}{a}_1+a_2+\cdots +a_m=n\\ a_1,a_2,\dots ,a_m\text{ are positive integers}\end{array}}{\min }\left(\frac{1}{a_1}+\frac{1}{a_2}+\cdots +\frac{1}{a_m}\right).$$

We prove the lemma by induction on $m$. For $m=1$, $f_m(n+m-1)=f_1(n)=\frac{1}{n}$, and the statement holds. Assume the statement holds for $m-1$, and consider the case with $m$ colors. For $i=1,2,\dots ,n$, let $x_i$ be the weight of the first color ball for person $P_i$, and assume without loss of generality that $x_1\le x_2\le \cdots \le x_n$. Choose non-negative integer $k$ such that $$x_1+x_2+\cdots +x_k\le A,x_1+x_2+\cdots +x_k+x_{k+1}>A.$$ Then $x_{k+1}\ge \frac{x_1+\cdots +x_{k+1}}{k+1}>\frac{A}{k+1}$. Therefore, for people $P_{k+1},P_{k+2},\dots ,P_n$ ($n-k$ people), the total weight of their $m-1$ balls (colors $2,3,\dots ,m$) is at most $$f_m(n+m-1)\cdot A-\frac{A}{k+1}\le f_{m-1}(n-k+m-2)\cdot A.$$ This inequality holds because $$\underset{b_1+\cdots +b_{m-1}=n-k+m-2}{\min }\left(\frac{1}{b_1}+\cdots +\frac{1}{b_{m-1}}\right)+\frac{1}{k+1}\ge \underset{a_1+\cdots +a_m=n+m-1}{\min }\left(\frac{1}{a_1}+\cdots +\frac{1}{a_m}\right).$$ By the induction hypothesis, we can select one ball (from colors $2,3,\dots ,m$) from each of $P_{k+1},P_{k+2},\dots ,P_n$ such that the total weight of each color among the selected balls is at most $A$. This completes the induction.

Substituting $A=\frac{1}{f_m(n+m-1)}$ in the lemma, we see that $C=\frac{1}{f_m(n+m-1)}$ satisfies the requirement. $$\text{Suppose }{f}_m(n+m-1)=\frac{1}{a_1}+\cdots +\frac{1}{a_m}\text{ where }{a}_1+\cdots +a_m=n+m-1.$$

If $C<\frac{1}{f_m(n+m-1)}$, consider the following scenario: For each person, the weight of their $k$-th color ball is $$\frac{1}{a_k}\times \frac{1}{\frac{1}{a_1}+\cdots +\frac{1}{a_m}}\left(>\frac{1}{a_k}\times C\right),$$ and the total weight of each person's $m$ balls is $1$. In this case, the number of selected balls of color $k$ is at most $a_k-1$, so the total number of selected balls would be at most $(a_1-1)+\cdots +(a_m-1)=n-1$, which is a contradiction. Therefore, the minimal $C$ is $C_{\min }=\frac{1}{f_m(n+m-1)}$. Let $n=mq+r$ where $1\le r\le m$, i.e., $n+m-1=m(q+1)+(r-1)=(r-1)(q+2)+(m+1-r)(q+1)$. Since $\frac{1}{x}$ is convex, the sum $\frac{1}{a_1}+\cdots +\frac{1}{a_m}$ is minimized when $a_1,\dots ,a_m$ consist of $(r-1)$ copies of $(q+2)$ and $(m+1-r)$ copies of $(q+1)$. Thus, $$C_{\min }=\frac{1}{f_m(n+m-1)}=\frac{1}{\frac{r-1}{q+2}+\frac{m+1-r}{q+1}}.$$ In particular, when $n=2025$ and $m=66$, the required $C$ is $\frac{248}{517}$. $\square$