Seven robbers divide the looted gold coins in such a way that, proceeding in alphabetical order, each takes as many as the sum of the digits in the number of gold coins not yet distributed. After two complete rounds, the gold runs out. Everyone got the same amount, except for the leader, who got more. What was the leader's position in the alphabetical order?
After the first robber takes as many gold bars as the sum of the digits in the number of the looted gold bars, the remaining number of bars will certainly be divisible by 9, since according to the rule of divisibility by 9, the remainder of the sum of the digits when divided by 9 is the same as the remainder of the number when divided by 9.

Another 13 times, a sum divisible by 9 is taken from a sum that is divisible by 9, always taking the sum of the digits of the amount, which is also divisible by 9, so a number divisible by 9 remains after each step.

Since the bars run out, the number of bars taken in the last step is a single-digit number, a number divisible by 9, because numbers with two or more digits are larger than the sum of their digits. Therefore, the number of bars taken in the 14th step is 9.

Since 14 times gold is taken and 13 of these times it is certainly a multiple of 9, the initial number is certainly greater than 100.

Let's consider the case when the number of gold bars is still three-digit, and a robber takes the sum of the digits, and the remaining number of gold bars is only two-digit:

$$
\overline{a b c}-(a+b+c)=99 a+9 b
$$

Since $a>0$, this number can only be two-digit if $a=1$ and $b=0$. Therefore, at some point during the takings, the number of gold bars will be 99.

Then the gold will decrease as follows:

| the number of bars | 99 | 81 | 72 | 63 | 54 | 45 | 36 | 27 | 18 | 9 | 0 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| the sum of the digits in the number of bars | 18 | 9 | 9 | 9 | 9 | 9 | 9 | 9 | 9 | 9 | 0 |

This is a total of 10 steps, so the 5th robber was the one who took 18 bars from 99. According to the text of the problem, everyone except the leader gets the same amount, so they each took 9 bars twice. Using this, the first five takings of bars are as follows:

| the number of bars | 135 | 126 | 117 | 108 | 99 | 81 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| the sum of the digits in the number of bars | 9 | 9 | 9 | 9 | 18 | 9 |

So the leader was the 5th in the list.

Based on the work of Markó Anna Erzsébet (Révkomárom, Selye János Gymnasium, 11th grade) and Vida Tamás (Győr, Kazinczy F. Gymnasium, 10th grade)