Let there exist 10 numbers that satisfy the condition. Clearly they're all nonnegative. Let $S,M,m$ be the sum, the maximum, the minimum of those numbers. Since $S-m\geq S-M \geq 0$, $(S-m)^2=m\geq(S-M)^2=M$. Here, since $M\geq m$, $M=m$ and thus all ten numbers are equal, which is a contradiction. Therefore there does not exist ten real numbers satisfying the condition.

Another solution, let there exist ten numbers $a_0,a_1,...,a_9$ satisfying the condition. Let $S$ be the sum of those. Here for all $i=0,1,...,9$ $(S-a_i)^2=a_i$ i.e. $a_i^2-(2S+1)a_i+S^2=0$. Let $p,q$ be the roots of $x^2-(2S+1)x+S^2=0$. Since not all of $a_i$ are equal, there must exist at least one $p$ and $q$ from $a_0,a_1,...,a_9$. Since $p,q,a_0,a_1,...,a_9$ must be nonnegative, $2S+1>S\geq p+q=2S+1$ (by Vieta's formula). Proof by contradiction.