Two intervals and a natural number What are the sufficient and necessary conditions on the real numbers $a,b,c$, and $d$ such that the same natural number $m$ is located in the same time in the two intervals $(a,b)$ and $(c,d)$. I remark that we can get $$max(a,c)<m<min(b,d)$$.

Clearly, one necessary condition is that $c < b,$ for otherwise there are no numbers
at all in common between the two intervals. Likewise $a < d.$
If "natural number" implies you are looking for a positive number,
there are additional conditions about the signs of $b$ and $d.$
But the conditions "$a < d$ and $c < b$" are satisfied
when $a < 3 < c < b < 4 < d$ (for example), in which case
there is no natural number that is in both intervals.
Since you are looking for something that does not require a value of $m$ to be
either presupposed or "guessed",
I think you must somehow explicitly produce a natural number 
from the values $a,$ $b,$ $c,$ and/or $d.$
Notice what a difference it makes if you write $c < \lfloor b\rfloor$ instead of $c < b,$
for example.
It is not possible that $3 < c < \lfloor b\rfloor < 4.$
It doesn't help much to write $\lfloor c\rfloor < b,$ because we can find $b$ and $c$
such that $3 = \lfloor c\rfloor < c < b < 4,$ 
but consider the implications of $\lceil c\rceil < b.$
It's possible to use these hints to tighten up the conditions for "some number in common"
to make conditions for "some natural number in common" if you don't get tripped up
by the borderline cases.
