Irrational

We represent the numbers like a vertices of a graph. We link to vertices, if the sum of them is rational.

Suppose there is an odd circle in the graph. Then if the numbers in this circle are $a_1, a_2, \dots a_{2n+1},$ then $a_1+a_2+\dots+a_{2n+1}=S$ is rational, because $(a_1+a_2)+(a_2+a_3)+\dots+(a_{2n+1}+a_1)=2S$ is rational. We can get $S$ in an other way: we add the circle's every second edge and the $2n+1$ . vertex (So $S=(a_1+a_2)+(a_3+a_4)+\dots+(a_{2n-1}+a_{2n})+a_{2n+1}$ ). However $(a_1+a_2), (a_3+a_4), \dots, (a_{2n-1}+a_{2n})$ are all rational, so $a_{2n+1}$ have to be rational too, but it is impossible.

So we get an even graph. In one group of vertices there are not less than $\frac{99}{2}$ points. So choosing these numbers the sum of in each pair is irrational.

So $50$ is the minimum? Yes, there is an example:

Let's have the $k+\sqrt{2}, l-\sqrt{2}$ numbers, where $k$ and $l$ are positive integers. If $k=l+1$ or $k=l-1,$ then at this puzzle the maximum is 50.