Given a positive integer $n(n \geqslant 2)$, find the largest $\lambda$ such that: if there are $n$ bags, each containing some balls of integer powers of 2 grams, and the total weight of the balls in each bag is the same, then there must be a certain weight of balls whose total number is at least $\lambda$. (The same bag can contain balls of equal weight.) (Supplied by Wang Jianwei)
Let's assume the heaviest ball weighs 1. First, we prove:
$$
\lambda_{\max } \geqslant\left[\frac{n}{2}\right]+1 \text {. }
$$

Let the total weight of the balls in each bag be $G$, then $G \geqslant 1$.
Assume the total number of balls of any weight is less than or equal to $\left[\frac{n}{2}\right]$. Considering the total weight of all balls in these $n$ bags, we get
$$
\begin{array}{l}
n \leqslant n G<\left[\frac{n}{2}\right] \cdot\left(1+2^{-1}+2^{-2}+\cdots\right) \\
=2\left[\frac{n}{2}\right] \leqslant 2 \cdot \frac{n}{2}=n .
\end{array}
$$

This is a contradiction.
Conversely, take a sufficiently large positive integer $s$, such that
$2-2^{-s} \geqslant \frac{2 n}{n+1}$.
Since $\left[\frac{n}{2}\right]+1 \geqslant \frac{n+1}{2}$, we have
$2-2^{-s} \geqslant \frac{2 n}{n+1} \geqslant \frac{n}{\left[\frac{n}{2}\right]+1}$.
Thus, $\left(\left[\frac{n}{2}\right]+1\right)\left(1+2^{-1}+\cdots+2^{-s}\right) \geqslant n \times 1$.
Therefore, we can take a continuous subsequence of sum 1 from
$$
\underbrace{1,1, \cdots, 1}_{\left(\left[\frac{n}{2}\right]+1\right) \uparrow}, \underbrace{2^{-1}, 2^{-1}, \cdots, 2^{-1}}_{\left(\left[\frac{n}{2}\right]+1\right) \uparrow}, \cdots, \underbrace{2^{-3}, 2^{-3}, \cdots, 2^{-3}}_{\left(\left[\frac{n}{2}\right]+1\right) \uparrow},
$$

from front to back, and we can do this at least $n$ times. Therefore,
$$
\lambda_{\max } \leqslant\left[\frac{n}{2}\right]+1 .
$$

In summary, $\lambda_{\max }=\left[\frac{n}{2}\right]+1$.