7. A square is cut into red and blue rectangles. The sum of the areas of the red rectangles is equal to the sum of the areas of the blue ones. For each blue rectangle, we write down the ratio of the length of its vertical side to the length of its horizontal side, and for each red rectangle - the ratio of the length of its horizontal side to the length of its vertical side. Find the smallest possible value of the sum of all the recorded numbers.
Answer: $1 / 2$. Example: two rectangles $1 / 2 \times 1$.

We can assume that the side of the original square is 1. The ratio of the sides of any rectangle is greater than or equal to its area, so each of the two sums ("red" and "blue") is not less than $1 / 2$. It remains to prove that one of these two sums is not less than 2.

Project the blue rectangles onto the vertical side of the square, and the red ones onto the horizontal side. In one of the two cases, the projections will cover the side completely (otherwise, there will be a point in the square not covered by any of the rectangles). Without loss of generality, let these be the vertical projections $a_{i}$ of the blue rectangles. Then $\sum a_{i} \geqslant 1$.

If $b_{i}$ are the horizontal sides of the blue rectangles, then $\sum a_{i} b_{i}=1 / 2$. By the Cauchy-Schwarz inequality, $\left(\sum a_{i} / b_{i}\right)\left(\sum a_{i} b_{i}\right) \geqslant\left(\sum a_{i}\right)^{2} \geqslant 1$, hence $\sum a_{i} / b_{i} \geqslant 2$, which is what we needed to prove.