Three. (25 points) Find the smallest positive integer $n$, such that among any $n$ irrational numbers, there are always three numbers, the sum of any two of which is irrational.
Three, take four irrational numbers $\{\sqrt{2}, \sqrt{3},-\sqrt{2},-\sqrt{3}\}$, obviously they do not satisfy the condition, hence $n \geqslant 5$.
Consider five irrational numbers $a, b, c, d, e$, viewed as five 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 $a+b$, $b+c$, and $c+a$ are all rational numbers. But $(a+b)+(c+a)-(b+c)=2a$, which contradicts the fact that $a$ is irrational.
(2) There must be a monochromatic triangle. Otherwise, there must be a red circle in the graph, i.e., each point is connected to two red lines and two blue lines (if there is a point $A$ connected to at least three lines of the same color $A B, A C, A D$, then $\triangle B C D$ is a monochromatic triangle of the other color, a contradiction). Assume the five numbers on the red circle are $a, b, c, d, e$ in sequence. Then $a+b, b+c, c+d, d+e, e+a$ are rational numbers.
$$
\text { By }(a+b)-(b+c)+(c+d)-(d+e)
$$
$+(e+a)=2a$ leads to a contradiction. Thus, the monochromatic triangle must be a blue triangle.
In summary, the minimum value of $n$ is 5.