Note that the number of squares that each kid gets is $B$ times a square number, where $B$ is the minimum number of congruent squares that a kid can cut one of the rectangles into. If a multiple of $B$ is prime, then either $B$ is prime and the other factor is 1 or vice versa, but this situation is impossible as there are multiple kids.

k.vasilev wrote: In a kindergarten, a nurse took $n$ congruent cardboard rectangles and gave them to $n$ kids, one per each. Each kid has cut its rectangle into congruent squares(the squares of different kids could be of different sizes). It turned out that the total number of the obtained squares is a prime number. Prove that all the initial squares were in fact squares. do you mean all initial rectangles were in fact squares in the last line?

Nice problem. For the sake of contrary say we have a $a\times b$ rectangle with $a<b$ . If $i^{\text{th}}$ kid has a square of side length $a_i$ after the process , we can rewrite $a=n_{i}(s)a_{i}$ and $b=t_{i}(s)a_{i}$ , where $n_{i}(s)$ and $t_{i}(s)$ are the number of squares on the sides of the original rectangle. Write $\frac{b}{a}=\frac{x}{y}$ where the last fraction is irreducible. Then $x\mid t_{i}(s)$ for all $i$ ,as $1<\frac{b}{a}=\frac{t_{i}(s)}{n_{i}(s)}$ so the total number of squares ($T$) is divisible by $x$ and $1<x<T$ , as $x\le t_{i}(s)$ , $T$ is composite , contradiction. $\blacksquare$