Example 8 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 equal (there can be balls of equal weight in the same bag), then there must be a certain weight of balls whose total number is at least $\lambda$. (2005 China National Team Selection Examination Problem)
Let the weight of the heaviest ball be 1, and let the total weight of the balls in each bag be $G$, then $G \geqslant 1$.

First, we prove that when $\lambda=\left[\frac{n}{2}\right]+1$ is valid, i.e., there must be a weight of balls whose total number is at least $\left[\frac{n}{2}\right]+1$.

Assume, for contradiction, that the total number of balls of any weight does not exceed $\left[\frac{n}{2}\right]$. Consider the total weight of all the balls in these $n$ bags, we have $n \leqslant n G<\left[\frac{n}{2}\right] \cdot\left(1+2^{-1}+2^{-2}+\cdots\right)=2\left[\frac{n}{2}\right] \leqslant 2 \cdot \frac{n}{2}=n$, a contradiction!

Next, we prove that $\lambda \leqslant\left[\frac{n}{2}\right]+1$. 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 $\left(\left[\frac{n}{2}\right]+1\right)\left(1+2^{-1}+\cdots+2^{-s}\right) \geqslant n \cdot 1$. Taking consecutive terms that sum to 1 from the beginning to the end, and at least $n$ times, so $\lambda \leqslant\left[\frac{n}{2}\right]+1$.

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