Let $m>1$ and $p_2,\cdots,p_k$ are all odd prime numbers less than $2m$.
$q,a_2,\cdots,a_k$ are arbitrarily selected integers(I mean no matter how you choose these numbers).
$$G_1=\lbrace n\in \mathbb Z|q < n \leq q+2m \rbrace,$$
$$G_i=\lbrace n\in \mathbb Z|n\not\equiv a_i \mod {p_i}\rbrace,$$
for $i=2,3,\cdots,k$.
$$G=\bigcap_{i=1}^k{G_i}.$$
Is it always true that #$|G|\geq 2$?
Especially,is it true that at least two of any consecutive $2m$ positive integers cannot be divided by odd prime numbers less than $2m$?
In other words, if you write $2m$ consecutive positive integers and cross out the multiples of $3, 5,\cdots,p_k$, then at least two of these $2m$ numbers will not be crossed out.
(1) When $m=2,k=2,p_2=3,$ four consecutive integers $q+1,q+2,q+3,q+4$,no matter what is $a_2$,we know that $G$ has at least two elements.
(2) For every $m>1$, we can choose $q$ and $a_i$ so that #$|G|=2$,for example, if
$$q=\frac{R-1}{2}-m,R=\prod_{i=2}^k{p_i},$$
$$a_i=0,i=2,\cdots,k,$$
then $G=\lbrace \frac{R-1}{2},\frac{R+1}{2} \rbrace.$
(3) The case $m=3,k=2,p_2=3,p_3=5.$
#$|G_1 \setminus G_2|=6/3=2,$ #$|G_1 \setminus G_3|\leq [6/5]+1=2,$
hence #$|G|\geq 2.$
(4) When $m\geq 36$, as $$\sum_{2<p<2m}{[\frac{2m}{p}]}>2m-2$$ maybe hold, I don't know if there's a counterexample.
Best Answer
To end this question, someone find a counterexample for $m=70$, where $$q\equiv -\lbrace 1,2,1,8,1,9,5,6,10,11,3,2,22,1,33,21,1,23,18,20,4,18,21,19,35,38,44,45,51,54,56,59,69 \rbrace \\ \mod {\lbrace 3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139 \rbrace }.$$
This means that $q\equiv -1\pmod 3,q\equiv -2\pmod 5,\cdots,q\equiv -69\pmod {139}.$
A solution is $q=264782491305295395386123607229983302927523861123269753$ and every number in $q+1,\cdots,q+140$ can be divided by some odd prime number less that $140$.