First show that each of $\sqrt{\frac{2005}{x+y}}$ etc. must be rational.

First, we will show that $\sqrt{\frac{2005}{x+y}}, \sqrt{\frac{2005}{y+z}}$, and $\sqrt{\frac{2005}{z+x}}$ must each be rational. For sake of simplicity, denote these radicands by $a, b$, and $c$. Let $d = \sqrt{a} + \sqrt{b} + \sqrt{c}$. Then $(d-\sqrt{a})^2 = (\sqrt{b} + \sqrt{c})^2$. Expanding, we get $d^2 - 2\sqrt{a} + a = b + 2\sqrt{bc} + c$. Rearranging gives $2\sqrt{bc} = d^2 + a - b - c - 2\sqrt{a}$, and squaring again gives $2bc = (d^2 + a - b - c)^2 - 4(d^2 + a - b - c)\sqrt{a} + 4a$. Hence we see that $\sqrt{a} = \sqrt{\frac{2005}{x+y}}$ must be rational. Similarly, we can obtain that $\sqrt{b} = \sqrt{\frac{2005}{y+z}}$ and $\sqrt{c} = \sqrt{\frac{2005}{z+x}}$ must also be rational. Therefore we may write $\sqrt{\frac{2005}{x+y}} = \frac{m}{n}$, where $m$ and $n$ are relative prime positive integers. Then $2005n^2 = (x+y)m^2$, so $m^2 | 2005n^2$. However, since $\gcd(m, n) = 1$, we must have $m^2 | 2005$. Since $2005 = 5 \cdot 401$ is squarefree, we must have $m = 1$. This gives $\sqrt{\frac{2005}{x+y}} = \frac{1}{n}$. In a similar fashion, we may do the same for the other two addends. Therefore, we may write $\sqrt{\frac{2005}{x+y}} + \sqrt{\frac{2005}{y+z}} + \sqrt{\frac{2005}{z+x}} = \frac{1}{n} + \frac{1}{p} + \frac{1}{q}$, for some integers $n, p, q$ (not necessarily primes - I'm just running out of letters to use). Since $\frac{1}{n} \le 1$, we may bound $1 \le \frac{1}{n} + \frac{1}{p} + \frac{1}{q} \le 3$ since our sum must be an integer. If $\frac{1}{n} + \frac{1}{p} + \frac{1}{q} = 3$, then $n = p = q = 1$, which yields no integer solutions for $x, y, z$. Now consider $\frac{1}{n} + \frac{1}{p} + \frac{1}{q} = 2$. By straightforward bounding arguments, we can obtain that $(n, p, q) = (1, 2, 2)$ or some permutation. It is easy to see that this also yields no integer solutions for $x, y, z$. Therefore we must have $\frac{1}{n} + \frac{1}{p} + \frac{1}{q} = 1$. Since our expression is symmetric, WLOG we may assume $n \le p \le q$. We can easily bound $1 < n \le 3$. If $n = 3$, then $n = p = q = 3$, which again yields no solutions for $x, y, z$. Hence $n = 2$.and $\frac{1}{p} + \frac{1}{q} = \frac{1}{2}$. This is a classical SFFT problem, for which we obtain $(p-2)(q-2) = 4$. The positive integer solutions to this are $p = q = 4$ and $(p, q) = (3, 6)$. In the former case, we obtain the solution $(x, y, z) = (4010, 4010, 28070)$. In the the latter case, there are no integer solutions for $x, y, z$. Therefore, the only unordered triple that satisfies the desired condition is $\boxed{(4010, 4010, 28070)}$, which is easily confirmed to work. I glossed over some of the equation solving, assuming people would know how to solve them. In case anyone needs help, here's the general technique. Once you obtain $n, p, q$, we have $\begin{cases} x+y = 2005n^2 \\ y + z = 2005p^2 \\ z + x = 2005q^2 \end{cases}$. The easiest way to solve this system is to first sum the equations and divide by $2$, to obtain $x+y+z = r$, for some constant $r$. If $r$ is not an integer then there are no integer solutions for $x, y, z$. Otherwise subtract each of the original equations from this equation to obtain $x, y, z$.