What is the greatest number of points in the plane such that the distance between any two of them is an odd integer? Suppose the origin is one of the points, we can take $(3,0)$ as the second point. If $(x,y)$ is some other point in the set I think we can use the fact that $x^2+y^2$, for $x$ and $y$ odd, is never an integer. There is also a lot of lines that the points certainly cannot belong to. But I could not go beyond.

The answer is $3$.
Assume the contrary, let's say we can find four points $O,A,B,C$ such that following six distances are all odd integers:
$$(a,b,c,a_1,b_1,c_1) = (BC,CA,AB,OA,OB,OC)$$
Since $O, A, B, C$ lies in the plane, they form a degenerate tetrahedron with volume $V = 0$. Express $V$ in terms of corresponding Cayley Menger determinant,
the distances satisfy
$$\left|\begin{matrix}
0 & 1 & 1 & 1 & 1\\
1 & 0 & a_1^2 & b_1^2 & c_1^2\\
1 & a_1^2 & 0 & c^2 & b^2 \\
1 & b_1^2 & c^2 & 0 & a^2\\
1 & c_1^2 & b^2 & a^2 & 0 \\
\end{matrix}\right| = 288V^2 = 0\tag{*1}$$
Recall if $u$ is an odd integer, then $u^2 \equiv 1 \pmod 8$.
Taking modulo $8$ on both sides of $(*1)$, we arrive at a contradiction.
$$4 = \left|\begin{matrix}
0 & 1 & 1 & 1 & 1\\
1 & 0 & 1 & 1 & 1\\
1 & 1 & 0 & 1 & 1 \\
1 & 1 & 1 & 0 & 1\\
1 & 1 & 1 & 1 & 0 \\
\end{matrix}\right| \equiv 0 \pmod 8$$
This means it is impossible to find four points in the plane whose pairwise distances are all odd integers.
Since it is trivial to find three points at odd integral distances from each other (eg. the vertices of an equilateral triangle of side $1$), the greatest number we seek is $3$.
