11.2. Given that the weight ratio of any two out of $N$ weights belongs to $\left[\frac{4}{5}, \frac{5}{4}\right]$, and these $N$ weights can be divided into 10 groups of equal weight, as well as 11 groups of equal weight. Find the minimum possible value of $N$.
11.2. $N_{\min }=50$.
An example for $N=50$.
In fact, 20 weights of $50 \mathrm{~g}$ and 30 weights of $40 \mathrm{~g}$ satisfy the requirement. These 50 weights can be divided into 10 equal groups, each consisting of 2 weights of $50 \mathrm{~g}$ and 3 weights of $40 \mathrm{~g}$. They can also be divided into 11 equal groups (5 groups each consisting of 4 weights of $50 \mathrm{~g}$, and 6 groups each consisting of 5 weights of $40 \mathrm{~g}$).
Now assume $N<50$.
Lemma: Suppose the ratio of the weights of any two weights is in $\left[\frac{4}{5}, \frac{5}{4}\right]$. If two groups of weights, one consisting of $k$ weights and the other of $l (k<l)$ weights, are equal in weight, then $k \geqslant 4$. When $k=4$, $l=5$ and each group consists of weights of the same weight.

Proof of the lemma: Let the weight of each group be $M$, and let $m$ be the weight of the lightest weight in the group of $l$ weights. Then the weight of each weight in both groups does not exceed $\frac{5}{4} m$. This gives us
$$
\frac{5}{4} k m \geqslant M \geqslant l m \geqslant(k+1) m,
$$
which implies $k \geqslant 4$.
If $k=4$, all inequalities in the above estimate become equalities, so $l=k+1=5$.

In the group of 5 weights, all weights are of weight $m=4 x$, and in the group of 4 weights, all weights are of weight $5 x$.
Returning to the original problem.
First, assume that among the 10 equal groups of weights, there are 2 groups with a different number of weights. Then, by the lemma and $N<50$, some groups consist of 4 weights each of weight $5 x$, and the remaining groups consist of 5 weights each of weight $4 x$. In this case, the weight of each group is $20 x$, so the total weight of the weights is $200 x$.

Since 200 is not a multiple of 11, these weights cannot be divided into 11 equal groups. This is a contradiction.

Therefore, the 10 equal groups of weights must have the same number of weights in each group. This implies $10 \mid N$.

If among the 11 equal groups, there are 2 groups with a different number of weights, then by the lemma, each group must have at least 4 weights, which implies $N \geqslant 44$. Since $N$ is a multiple of 10, we get $N \geqslant 50$, a contradiction.

If among the 11 equal groups, each group has the same number of weights, then $11 \mid N$, which implies $110 \mid N$. This contradicts the assumption $N<50$.