Answer: 16. Solution: Example. Let's divide all 24 cells into eight triples, where each triple consists of three cells adjacent to one vertex of the cube. Any two cells in the same triple share a common side. Since the number of marked cells is the same as the number of triples, there must be exactly one marked cell in each triple. Place 16 detectors such that there are two detectors in each triple. If one of the detectors in a given triple triggers, we have found the marked cell in that triple; if neither detector triggers, the marked cell is the one without a detector. Evaluation: Suppose we place fewer than 16 detectors. Then there will be a triple with at least two cells without detectors (let's call them "free cells")—mark these cells on the unfolded cube diagram with a dark background. On the same unfolded diagram, mark 7 cells with the letter A as shown in the figure. Now note that if we mark the seven cells A and one of the free cells with invisible ink, the detectors will not allow us to determine which of the free cells is marked. Therefore, it is not possible to get by with fewer than 16 detectors.

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4. There is a cube, each face of which is divided into 4 identical square cells. Oleg wants to mark 8 cells with invisible ink so that no two marked cells share a side. Rustem has detectors. If a detector is placed in a cell, the ink on it becomes visible. What is the minimum number of detectors Rustem can place in the cells so that, no matter which cells Oleg marks afterward, all the marked cells can be identified? (R. Zhinodarov, O. Dmitriev)