7.1. Is it possible to place rational numbers at the vertices of a square $A B C D$ such that the sum of the numbers at vertices $A$ and $B$ is 1, at vertices $B$ and $C$ is 2, at vertices $C$ and $D$ is 3, and at vertices $A$ and $D$ is 4?
Answer: No.

Solution. Let the number $x$ be placed at vertex $A$, then the number $1-x$ will be placed at vertex $B$, the number $1+x$ at vertex $C$, and the number $2-x$ at vertex $D$. Then, for any value of $x$, the sums of the numbers on the three sides will be $1, 2, 3$, respectively. For the sum of the numbers on side $AD$ to be equal to 4, we need to find $x$ from the condition $x+(2-x)=4$. However, such a number does not exist.