Does there exists three intervals $I_1,I_2,I_3$ each of length as an even integer such that $I_1 \cap I_2$ , $I_2 \cap I_3$ and $I_3 \cap I_1$ are of length as an odd integer?
Tried few examples and observed that it's difficult to satisfy all intersections having odd length. Maybe we should prove it is not possible.
On
No. Name the endpoints so $I_1=[a_1,b_1]$, $I_2=[a_2,b_2]$ and $I_3=[A_3,b_3]$, and assume that the intervals are order by their lower endpoint, i.e. $a_1\leq a_2\leq a_3$. It's also clear that no two of the intervals can be contained within each other as that would make their intersection equal to the smaller and of even length.
If $a_1\neq 0$ we can translate all the intervals so it is. As $I_1$ has even length, that means $b_1$ is even. As $I_1\cap I_2$ has uneven length, we see that $a_2$ must be odd (the intersection has length $b_1-a_2$). A similar argument between $I_1$ and $I_3$ proves that $a_3$ must also be odd, but a similar argument between $I_2$ and $I_3$ shows that $a_3$ must be even.
You can assume that the intervals have integer endpoints and are closed. Write them as $I_j=[a_j,b_j]$ with $a_1\le a_2\le a_3$. If $b_j\ge b_k$ with $k>j$ then $I_k\subseteq I_j$ which is impossible. So $b_1<b_2<b_3$. Then $I_1\cap I_2=[a_2,b_1]$, $I_1\cap I_3=[a_3,b_1]$ and $I_2\cap I_3=[a_3,b_2]$. Then $a_2\equiv b_2\equiv a_3+1\equiv b_1 \equiv a_2+1\pmod2$.