Twelve chairs are arranged in a row. Sometimes a person sits on one of the free chairs. At this point, exactly one of his neighbors (if they were present) stands up and leaves. What is the maximum number of people that can be sitting at the same time, if initially all the chairs were empty?

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Evaluation. It is impossible for all chairs to be occupied simultaneously, because at the moment when a person sits on the last unoccupied chair, one of his neighbors will stand up. Therefore, the number of people sitting simultaneously cannot exceed 11.

Example. Let's show how to seat 11 people. Number the chairs from 1 to 12. The first chair is easy to occupy. The second chair will be occupied in two stages. In the first stage, a person sits on the third chair, and in the second stage, we seat a person on the second chair, and the person sitting on the third chair stands up. We proceed similarly: if the chairs numbered from 1 to \( k \) are occupied, we first seat a person on the chair numbered \( k+2 \), and then seat a person on the chair numbered \( k+1 \), thereby freeing the chair numbered \( k+2 \). After this operation is performed for all \( k \) from 1 to 10, the chairs numbered from 1 to 11 will be occupied, and the twelfth chair will be free.

## Answer

11.