2 and 3 Imply 4

Let's prove this by contradiction.

Suppose that it is possible to fill all boxes with positive values. Let $A, B, C, D, E, F, G, H$ be distinct positive integers, such that $\Large A \div B \div C \div D = E \div F \div G \div H$ Multiplying both sides by the common denominators, we obtain $\Large A\times F \times G \times H = B \times C \times D \times E$ Note that $9!$ has $9$ distinct positive integer factors from $1$ to $9$ . Looking at the factors, we see that there are $2$ integers (5 and 7) that are all relatively prime with other $7$ integers. But since we are filling all $8$ boxes with distinct integers, either $5$ or $7$ is filled, contradiction.

Thus, there is $\boxed{\text{no}}$ way to fill all boxes with positive integers.

Note: This illustrates pigeonhole principle