$x_{1}-x_{2}=1$

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Let $x_{1}+x_{2}+\ldots+x_{n}=S$. Squaring the equation $\sqrt{x_{i}}+\sqrt{S-x_{i}}=\sqrt{x_{j}}+\sqrt{S-x_{j}}$ and combining like terms, we get

$x_{i}\left(S-x_{i}\right)=x_{j}\left(S-x_{j}\right) \quad \Leftrightarrow \quad\left(x_{i}-x_{j}\right)\left(x_{i}+x_{j}-S\right)=0$.

Thus, for each pair of unknowns, either $x_{i}=x_{j}$ or $x_{i}+x_{j}=S$.

$x_{1}>x_{2}$, so $x_{1}+x_{2}=S$. Since all unknowns are non-negative, $x_{3}=x_{4}=\ldots=x_{n}=0$.

$x_{2}+x_{3}=x_{2}<S$, therefore, $x_{2}=x_{3}=0$.

## Answer

$(1,0, \ldots, 0)$