Suppose that such numbers exist. Since they are equal to some squares, each of these numbers is nonnegative. Let's denote the sum of all these 10 numbers by $S$. Let's pick one of these numbers and denote it by $a$, then $S \ge a$, and also
$$a = (S - a)^2 \Rightarrow a^2 - (2S + 1)a + S^2 = 0.$$
If there are two different $a$, satisfying this condition, then by Vieta's theorem their product is equal to $S^2$. However, this means that one of them is greater than $S$, which contradicts what we proved above. Thus, there is only one possible $a$, and it follows that all numbers are equal.