Three, (25 points) Find the smallest positive integer $n$, such that among any $n$ irrational numbers, there are always 3 numbers, where the sum of any two of them is still irrational.


Three, take 4 irrational numbers $\{\sqrt{2}, \sqrt{3},-\sqrt{2},-\sqrt{3}\}$, clearly they do not satisfy the condition, hence $n \geqslant 5$.

Consider 5 irrational numbers $a, b, c, d, e$. View them as 5 points. If the sum of two numbers is a rational number, then connect the corresponding two points with a red line; otherwise, connect them with a blue line.
(1) There is no red triangle. Otherwise, assume without loss of generality that $a+b, b+c, c+a$ are all rational numbers. Since $(a+b)+(c+a)-(b+c)=2a$, this contradicts the fact that $a$ is irrational.
(2) There must be a monochromatic triangle. Otherwise, the graph must contain a red cycle, with the 5 numbers at the vertices, where the sum of any two is a rational number. Suppose $a+b, b+c, c+d, d+e, e+a$ are rational numbers, then by $(a+b)-(b+c)+(c+d)-(d+e)+(e+a)=2a$ we derive a contradiction. Thus, the monochromatic triangle must be a blue triangle. Therefore, the minimum value of $n$ is 5.