Squaring the first chain of equalities, and subtracting the common sum, then squaring again, we get $x_i(S-x_i)$ is constant as $i$ varies, where $S$ is the total sum of the $x_i$. if this constant value is $C$, then $x_i^2-Sx_i+C=0$. This is a quadratic, so there are at most two values of $x$ s.t. $x^2-Sx+C=0$. This implies that the $x_i$ take on at most $2$ different values. Let these two values be $a$ and $b$. Since $x_1-x_2=1$, if $x_2$ is WLOG $a$, then $x_1=a+1$, so therefore $b=a+1$. Since the vertex of the quadratic is at $x=\frac{S}{2}$, and the two roots are symmetric about this vertex, we must have $\frac{S}{2}=a+\frac{1}{2}$, so $S=2a+1$. Since $x_1+x_2=a+1+a=2a+1=S$, we must have $x_3+x_4+\cdots+x_n=0$. Since each $x_i$ is either $a$ or $a+1$, and both of these are non-negative to ensure that all of the square roots in the original question are well-defined, we must have them all equal to $0$. Therefore, we must have $x_3=x_4=\cdots=x_n=a=0$, so every $x_i$ is $0$ except $x_1$, which is $a+1=0+1=1$. It is easy to check that this works. Cheers, Rofler

Let be $S=x_1+x_2+\ldots+x_n$, we have $\sqrt x_1+\sqrt{S-x_1} \stackrel{(1)}{=} \sqrt x_2+\sqrt{S-x_2}\Longleftrightarrow$ $\frac{x_1-x_2}{\sqrt{x_1}+\sqrt{x_2}}=\frac{x_1-x_2}{\sqrt{S-x_1}+\sqrt{S-x_2}}\Longleftrightarrow$ $\sqrt x_1+\sqrt x_2=\sqrt{S-x_1}+\sqrt{S-x_2}\Longleftrightarrow \sqrt x_1-\sqrt{S-x_1}\stackrel{(2)}{=}\sqrt{S-x_2}-\sqrt x_2$ From $(1) \cdot (2)\Longrightarrow$ $2x_1-S=S-2x_2\Longleftrightarrow$ $S=1\Longleftrightarrow x_1+x_2+\ldots+x_n=x_1-x_2\Longleftrightarrow 2x_2+x_3+\ldots+x_n=0$ so $x_1=1, x_2=x_3=\ldots=x_n=0$.