9.5. In the cells of an 8 by 8 board, tokens are placed such that for each token, the row or column of the board in which it lies contains no more than three tokens. What is the maximum possible number of tokens on the board?
Answer: 30.

Solution: By swapping the verticals and horizontals, we assume that only the left $x$ verticals and the bottom $y$ horizontals contain more than 3 chips. From the condition, it follows that there are no chips at all in the lower left rectangle at the intersection of these verticals and horizontals. Each vertical passing through the chips in these horizontals contains no more than three chips, and each horizontal passing through the chips in these verticals contains no more than three chips. Then the total number of chips on the board does not exceed $3(8-x)+3(8-y)=48-3(x+y)$, which does not exceed 30 when $x+y \geq 6$.

Let $x+y \leq 5$ and $x \leq y$, then $x \leq 2$. We estimate the total number of chips on the board from above as $3(8-x)+x(8-y)=24+5x-xy$. For $x=2$, this is no more than 30, for $x=1, x=0$ - no more than 29 and 24, respectively.

Example of placing 30 chips: all cells of the three left verticals and the three bottom horizontals are filled, except for the lower left 3 by 3 square. Evaluation: 5 points for the estimate. 2 points for the example. Any incorrect estimate with any reasoning - 0 points.