Assume without loss of generality that the weight of the heaviest ball is 1. We first prove that $\lambda_{\max } \geqslant\left[\frac{n}{2}\right]+1$.
Let the total weight of the balls in each bag be $G$, then $G \geqslant 1$. Suppose the total number of balls of any weight does not exceed $\left[\frac{n}{2}\right]$, then by considering the total weight of all balls in $n$ bags from two aspects, we get
$$
n \leqslant n G \leqslant\left[\frac{n}{2}\right]\left(1+\frac{1}{2}+\frac{1}{2^{2}}+\cdots\right)<2\left[\frac{n}{2}\right] \leqslant n,
$$

which is a contradiction. Therefore, the assertion holds.
On the other hand, take a sufficiently large integer $k$, such that $2-2^{-k} \geqslant \frac{2 n}{n+1}$, then since $\left[\frac{n}{2}\right]+1 \geqslant \frac{n+1}{2}$, we have
$$
2-2^{-k} \geqslant \frac{2 n}{n+1} \geqslant \frac{n}{\left[\frac{n}{2}\right]+1},
$$

thus
$$
\left(\left[\frac{n}{2}\right]+1\right)\left(1+\frac{1}{2}+\frac{1}{2^{2}}+\cdots+\frac{1}{2^{k}}\right) \geqslant n \times 1 .
$$

Therefore, in
$$
\underbrace{1,1, \cdots, 1}_{\left(\left[\frac{n}{2}\right]+1\right) \uparrow}, \underbrace{\frac{1}{2}, \frac{1}{2}, \cdots, \frac{1}{2}}_{\left(\left[\frac{n}{2}\right]+1\right) \uparrow}, \cdots, \underbrace{\frac{1}{2^{k}}, \frac{1}{2^{k}}, \cdots, \frac{1}{2^{k}}}_{\left(\left[\frac{n}{2}\right]+1\right) \uparrow}
$$

we can sequentially take a continuous subsequence with a sum of 1 from the front to the back, and we can take at least $n$ times. Hence $\lambda_{\text {max }} \leqslant\left[\frac{n}{2}\right]+1$. Combining the two aspects, we get $\lambda_{\max } \leqslant\left[\frac{n}{2}\right]+1$.
223:1 Hyun 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 in weight, and the total weight of the balls in each bag is equal, 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.)