Let $n$ be a positive integer, and there are $2^{n}$ weights with distinct masses. Each weighing involves dividing the current weights into two groups of equal number and placing them on a balance to determine which group is heavier, then keeping the heavier group (if the two groups have the same mass, either group can be kept). After $n$ weighings, one weight remains. Question: What is the minimum rank, from heaviest to lightest, of this weight among the original $2^{n}$ weights?
Let the masses of the weights be $a_{1}<a_{2}<\cdots<a_{2^{n}}$.
We call the heavier group in each weighing the "winning group" and the lighter group the "losing group".

For $1 \leqslant k \leqslant n$, let the average mass of the winning group in the $k$-th weighing be $b_{k}$, and the mass of the last remaining weight be $a^{\prime}$.

Since the mass of the winning group in each weighing is no less than that of the losing group, we have:
$b_{1} \leqslant b_{2} \leqslant \cdots \leqslant b_{n-1} \leqslant b_{n}=a^{\prime}$.
Thus, for $1 \leqslant k \leqslant n$, we have
$a^{\prime} \geqslant b_{n} \geqslant$ the average mass of the losing group in the $k$-th weighing
$\geqslant$ the minimum mass in the losing group of the $k$-th weighing.
Since $a_{1}, a_{2}, \cdots, a_{2^{n}}$ are all distinct, $a^{\prime}$ is greater than the minimum masses of the $n$ losing groups. Therefore, from heaviest to lightest, $a^{\prime}$ is at least the $2^{n}-n$-th.
Now, let's construct the solution.
By the invariance under translation, it doesn't matter if $a_{1}, a_{2}, \cdots, a_{2^{n}}$ include negative numbers. For $1 \leqslant i \leqslant 2^{n}$, let $a_{i}=-2^{2 n-i}$.
Notice that, $a_{i}<a_{i+1}+a_{i+2}+\cdots+a_{2 n}$.
Thus, the group containing the smallest remaining mass in each weighing must be the losing group. For $1 \leqslant k \leqslant n$, in the $k$-th weighing, let $a_{k}$ and the $2^{n-k}-1$ heaviest weights at that time be in the same group, then they will be the losing group. This way, only $a_{n+1}$ remains at the end.
In conclusion, the minimum value sought is $2^{n}-n$.