Let's generalize the problem by replacing $2025$ with $n$ and $66$ with $m$. For any positive integers $m \le n$, define:
$$f_m(n) = \min_{\substack{a_1+a_2+\dots+a_m=n \\ a_1, a_2, \dots, a_m \text{ are positive integers}}} \left( \frac{1}{a_1} + \frac{1}{a_2} + \dots + \frac{1}{a_m} \right).$$

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 $x_1 \le x_2 \le \dots \le x_n$. Choose non-negative integer $k$ such that
$$x_1 + x_2 + \dots + x_k \le A, \quad x_1 + x_2 + \dots + x_k + x_{k+1} > A.$$
Then $x_{k+1} \ge \frac{x_1+\dots+x_{k+1}}{k+1} > \frac{A}{k+1}$. 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
$$\min_{b_1+\cdots+b_{m-1}=n-k+m-2} \left( \frac{1}{b_1} + \cdots + \frac{1}{b_{m-1}} \right) + \frac{1}{k+1} \ge \min_{a_1+\cdots+a_m=n+m-1} \left( \frac{1}{a_1} + \cdots + \frac{1}{a_m} \right).$$
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
$$\frac{1}{a_k} \times \frac{1}{\frac{1}{a_1} + \dots + \frac{1}{a_m}} \left( > \frac{1}{a_k} \times C \right),$$
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 $(a_1 - 1) + \dots + (a_m - 1) = n - 1$, 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 $\frac{1}{a_1} + \dots + \frac{1}{a_m}$ 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 $\frac{248}{517}$.
$\square$