We have $\left(a-\frac 12\right)^2+\left(b-\frac 12\right)^2+\left(c-\frac 12\right)^2+\left(d-\frac 12\right)^2-1=0$. Hence, $0\ge \left(a-\frac 12\right)^2-1\Rightarrow \frac 32\ge a$. Similarly, $\frac 32\ge a,b,c,d$. Thus, every real number $x\ge \frac 32$ satisfies the condition. See that none of the numbers in the interval $\left(\frac 12, \frac 32\right)$ works because $\left(\frac 12,\frac 12,\frac 12,\frac 32\right)$ is balanced. Also, none of the numbers in the interval $(0, 1)$ works because $(0, 0, 0, 1)$ is balanced. Thus, our answer is all real numbers greater than or equal to $\frac 32$.

Solution that I found during contest. Take $b=c=d=0, a=1$ $\implies$ $(x-1)x^3\geq 0$ $\implies$ $x\geq 1$. Take $b=c=d=\frac{1}{2}, a=\frac{3}{2}$ $\implies$ $(x-\frac{3}{2})(x-\frac{1}{2})^3\geq 0$ and since $x\geq 1>\frac{1}{2}$ we get $x\geq \frac{3}{2}$. Since $\sum (2a-1)^2=4$ we get that $a,b,c,d\leq \frac{3}{2}$. Since $x\geq \frac{3}{2}$ we get $x\geq a,b,c,d$ so all $x\geq \frac{3}{2}$ works. So answer is all real $x\geq \frac{3}{2}$.