Obviously, it is possible to rearrange the numbers in each row so that they increase. Denote the initial row as $0^{\text{th}}$, the next - $1^{\text{st}}$ and so on.

Let us describe the properties of the numbers inside rows beginning from the $1^{\text{st}}$ one.

Lemma 1. If $n$ occur in $k$ th row, then the amount of numbers "n" is divisible by $n$.

Proof. Number $n$ occurs in the $1^{\text{st}}$ row if there are $n$ equal numbers in the $0^{\text{th}}$ row, say $m$. Then under these $n$ numbers "$m$" one writes a group of numbers $n$. If another number $l$ in the $0^{\text{th}}$ row occurs $n$ times too, then a new group which contains $n$ numbers equal to $n$ will emerge in the $1^{\text{st}}$ row. The number $n$ may occur in the rows after $1^{\text{st}}$ one if there were numbers $n$ in the previous row and one wrote $n$ under them once again or there was a group of exactly $n$ equal numbers different from $n$. Hence, their amount is divisible by $n$.

Lemma 2. If there is a group of numbers "n" in the $k$ th row then the same number will be written under these numbers but this number is not obliged to be equal to $n$.

Proof. It is an easy consequence of the property that if in the previous row there were two equal numbers then under them one should write the same numbers too.

Consider the first row. Let $l$ be the smallest number inside it. That means the amount of the numbers "$l$" will be equal to $sl$, $s \ge 1$. The other numbers are greater than $l$.

If $l$ occur exactly $l$ times, then under them only one can write only numbers $l$. They cannot decrease since there are exactly $l$ of them and they cannot increase because in this case we need $l$ equal numbers but there is no such group. So we can erase these numbers and consider a row with $n-l$ numbers. Now we can look for the smallest number in the new row.

If there is a group of $l$ consisting of $sl$ numbers, $s > 1$, then in the next group one has to write $sl$ under them. Consider some $m$ in the 1st row such that: $l < m < 2l$. Analogously, if there are exactly $m$ of them then they cannot change in the lower rows. The numbers which are less then $m$ either do not change (and does not impact on the changes of the numbers $m$) or at least two times increase, so $m$ cannot occur under them. So we can ignore such numbers. If there are $tm$ of them then they are replaced by $tm$ in the consecutive rows.

Hence in the 2nd row we need only consider the numbers which are greater or equal to $2l$.

Analogously, if the least number in the 2nd row is $l_2$ (among the numbers which we need to consider), then in the 3rd row the least such number will be at least $2l_2$ and so on.

It is easy to see that a number greater then $n$ cannot emerge in the rows. So the greatest possible number of steps is $k$, for which $2^k \le n$. But we have to count the move from 0th to 1st row. So at most we can have $k+1$ steps.

Let the 0th row has the following appearance:
$$(n-1), (1), (2, 2), (4, 4, 4), \dots, \underbrace{(2^{k-1}, 2^{k-1}, \dots, 2^{k-1})}_{2^{k-1}}, \underbrace{(m, m, \dots, m)}_{m} \text{ (0th row)}$$

$$(1, 1), (2, 2), (4, 4, 4), \dots, \underbrace{(2^{k-1}, 2^{k-1}, \dots, 2^{k-1})}_{2^{k-1}}, \underbrace{(m, m, \dots, m)}_{m} \text{ (1st row)}$$
$$(2, 2, 2, 2), (4, 4, 4, 4), \dots, \underbrace{(2^{k-1}, 2^{k-1}, \dots, 2^{k-1})}_{2^{k-1}}, \underbrace{(m, m, \dots, m)}_{m} \text{ (2nd row)}$$
$$(4, 4, 4, 4, 4, 4, 4), \dots, \underbrace{(2^{k-1}, 2^{k-1}, \dots, 2^{k-1})}_{2^{k-1}}, \underbrace{(m, m, \dots, m)}_{m} \text{ (3rd row)}$$
$$\text{(2 k-1 , 2 k-1 , , 2 k-1 ) 2 k , (m, m, , m) m (k-1 th row)}$$
$$\text{(2 k, 2 k, , 2 k) 2 k , (m, m, , m) m (k th row)}$$

Now let us understand why the condition $m = 2^k - n = 2^i$ is important. If there is a group consisting of $2^i$ then there is a group of the numbers $2^i$, which will give a group of $2^{i+1}$ in the next row. So in the described process the numbers change as: $2^0 \to 2^1 \to 2^2 \to \dots \to 2^i$, in the next step the numbers will not change and the changes will last shorter.

For instance if there are 5 numbers in the beginning then we will have the following estimates: $2^2 = 4 < 5$, so the answer is at most 3 steps. But if we consider the 0th row as 1; 4; 2; 2; 5 we will get the following 1st row: 1; 1; 2; 2; 1, 2nd row: 3; 3; 2; 2; 3. The same situation is if $n = 2^s + 1$.

Analogously if $n = 2^s + 2$, because in this case either a group of two equal numbers will emerge or there will be two distinct numbers in the 0th row. The last case when we cannot increase the amount of numbers - $n = 2^s + 4$. Here we have that if 4 decomposed into several integer summands then one of this summands would be equal to some power of 2. In this case there will emerge a group of three one's in the first row. So we do not need $2^k \le n$, but $3 \cdot 2^{k-1} \le n$. Now we are going to determine for which $n$ it is possible. $3 \cdot 2^{k-1} \le n = 2^k + 4$. Then $2^{k-1} \le 4$, or $k \le 3$. So for $n=5$, $n=6$ and $n=12$. For $n=5$ we have already seen that it is impossible, for $n=6$ we have the following situation in 0th row: $1; 2; 2; 2; 5; 6 \rightarrow 1; 3; 3; 3; 1; 1 \rightarrow 3; 3; 3; 3; 3; 3 \rightarrow 6; 6; 6; 6; 6; 6$, Analogously if $n=12$.

In case of $2^i \ge 8$ it suffices to take instead of $(\underbrace{m, m, ..., m}_{m})$ in the basic example two groups: $(\underbrace{s, s, ..., s}_{s})$ and $(\underbrace{t, t, ..., t}_{t})$, where $s = 2^{i-1}-1$ and $t = 2^{i-1}+1$, which will not change during the whole procedure and will not interfere with the basic example.