The required maximum is $4$, an example of a special $4$-element set being $M = \{\sqrt{2} - 1, \sqrt{2} + 1, 2 - \sqrt{2}, -2 - \sqrt{2}\}$.

We will prove that a special set cannot have more than $4$ elements. Obviously, the second condition implies that all the elements of a special set are irrational. We will use the following remarks.

$\mathbb{R}_1$. If $x, y, z$ are three distinct elements of $M$, then $x+y$, $x+z$ and $y+z$ cannot be all rational.

If we assume the opposite, then $2(x + y + z) \in \mathbb{Q}$, whence $x+y+z \in \mathbb{Q}$, which, in turn, leads to $x \in \mathbb{Q}$, false.

$R_2$. If $x, y, z$ are three distinct elements of $M$, then $xy$, $xz$ and $yz$ cannot be all rational.

If we assume the opposite, then $x^2yz = (xy) \cdot (xz) \in \mathbb{Q}$ and $yz \in \mathbb{Q}^*$ lead to $x^2 \in \mathbb{Q}$, false.

$R_3$. If $x, y \in M$ and $xy \in \mathbb{Q}$ then, for every $z \in M$, $x + z \in \mathbb{Q}$ and $y + z \in \mathbb{Q}$.

If the opposite happens, then the assumption, $R_1$ and $R_2$ imply $x + z \in \mathbb{Q}$ and $yz \in \mathbb{Q}$, or $y + z \in \mathbb{Q}$ and $xz \in \mathbb{Q}$. In the first case, from $xy \in \mathbb{Q}$ and $yz \in \mathbb{Q}$ follows $xy + yz = y(x + z) \in \mathbb{Q}$ and, since $x + z \in \mathbb{Q}$ and $x + z \neq 0$, $y \in \mathbb{Q}$, false. The second case is similar.

Suppose now that there exists a special set with at least five elements $a, b, c, d, e$. Remark $R_1$ shows that at least two elements have an irrational sum – let them be $a$ and $b$. Then $ab \in \mathbb{Q}$ and $R_3$ implies that $a + c$, $a + d$, $a + e$ are rational. According to $R_1$, numbers $c+d$, $c+e$ and $d+e$ cannot be rational, and the assumption implies that $cd$, $ce$ and $de$ are rational, which contradicts $R_2$.