Help me please to solve the following problem.

There are $n$ arithmetic progressions of the form:

$$(2i+1)k + x_i,~~~~ i = 1,\ldots,n, k \geq 0$$

Initial integer terms $x_i \geq 0$ are varying.

The problem is to cover $m(n)$ - the maximum possible number of consecutive integers starting with $ 1 $ (the number is covered, if it belongs to at least one of these progressions).

For example, is it true, that: $$m(n) \approx \operatorname*{LCM}\limits_{i=1,\,\ldots\,,\,n}(\{2i+1\}) \text{?}$$

But it is so much and I am in search of the best solutions for this problem.

Numerical simulation shows something like this: $m(n) \approx Cn$

**Update**: Thank you for your answers! I did not imagine that it is so difficult problem.

**Question**: What is known about bounds of $m(n)$?

Is it proposition about upper bound whether true or not:

For fixed integer $k \geq 0$, and for all integers $x_i$ perform:

If $$u = \operatorname*{LCM}\limits_{i=1,\,\ldots\,,\,n}(\{(2i+1)k+x_i\})+1$$

Then number $u$ is not belong to any of $n$ arithmetic progressions of the form: $(2i+1)k+x_i,~~~i = 1,\ldots,n$ ?

It seems to me that it is obviously true.