11.3. In each row of the $100 \times n$ table, the numbers from 1 to 100 are arranged in some order, with no repetitions within a row (the table has $n$ rows and 100 columns). It is allowed to swap two numbers in a row that differ by 1, provided they are not adjacent. It turns out that it is impossible to obtain two identical rows using such operations. For what largest $n$ is this possible?

(M. Antipov)
Answer: $2^{99}$.

Solution. We will associate a string $x_{1}, x_{2}, \ldots, x_{100}$ of numbers from 1 to 100 with a sequence of 99 signs $$ according to how the adjacent numbers are ordered. That is, if $x_{k} < x_{k+1}$, the $k$-th sign is $<$, and if $x_{k} > x_{k+1}$, the $k$-th sign is $>$. Note that the allowed operations on the string do not change the corresponding sequence of signs. Indeed, in the pair of numbers $x_{k}$ and $x_{k+1}$, no more than one number changes and by no more than 1. Therefore, the inequality sign between them cannot change to the opposite. We will associate each permutation of signs with a placement of numbers according to the following rule (we will call such placements selected). For $k=1,2, \ldots, 99$, if $x_{k} > x_{k+1}$ (i.e., the $k$-th sign is $<$), we place the largest of the previously unselected numbers at the position of $x_{k}$, and if $x_{k} < x_{k+1}$, we place the smallest of the previously unselected numbers at the position of $x_{k}$. Note that in this sequence of operations, the numbers not selected in the first $k$ steps will form a segment of the natural number sequence (and the selected ones will be several of the largest and several of the smallest numbers from 1 to 100). In particular, $x_{1}=1$ or $x_{1}=100$. The number $x_{100}$ will be replaced by the only remaining number.

It is easy to see that the resulting placement of numbers corresponds to the selected sequence of signs. The total number of selected placements will be the same as the number of different sequences of signs, i.e., $2^{99}$. As stated above, no two selected strings can be made identical by allowed operations. Thus, by filling a $100 \times 2^{99}$ table with selected strings, we obtain an example for $n=2^{99}$.

Now we will prove that from any string $A$ of length 100, we can obtain a selected string $B$ with the same sequence of signs. This will imply that $n \leqslant 2^{99}$. Suppose the first $t-1$ signs of the given string $A=\left(x_{1}, \ldots, x_{100}\right)$ and the selected string $B=\left(y_{1}, \ldots, y_{100}\right)$ coincide. If $t=100$, then the strings themselves coincide. Let $t<100$. Without loss of generality, assume that $x_{t}<x_{t+1}$. As stated above, the sets of numbers $y_{t}, \ldots, y_{100}$ and $x_{t}, \ldots, x_{100}$ coincide and form a segment of the natural number sequence with the smallest number $x_{t}$. Since $y_{t}<y_{t+1}$, we can change the number at position $t$ in string $A$ to a number one less, until the number at position $t$ is $x_{t}$. Thus, we have achieved the coincidence of the first $t$ symbols of our string with the selected string $B$. Therefore, such operations can transform any string into a selected string, which completes the proof of the estimate.

Comment. Points for all the specified advances are not cumulative.

Only the answer - 0 points.

Transition to inverse permutations - 0 points.

The correct set of representatives is indicated - 1 point.

The correct set of representatives is indicated and it is proven that from any permutation one can transition to some representative. It is not proven that one cannot transition between representatives - 3 points.

It is proven (mainly formulated) that the sequence of signs of differences between adjacent numbers is preserved - 2 points.

It is proven that $n \leqslant 2^{99}$ - 3 points.

Incomplete induction - 0 points.