Suppose i have $n$ positive $q_i>1, i\in\{1,2\dots n\}$ integers. The multiples of these $q_i>1$ integers form sequences on number line with length $l\geqslant1$. My question is: Is it possible, for given $q_i>1$ integers to calculate maximum possible value for $l$? I do not want to use "brute-force" type approach. What would be most efficient way doing it?
If it simplifies the problem, one can also assume that $q_i>1$ integers are coprime.
For example: Suppose $n=2, q_1=2, q_2=3 $ the sequences of multiples (I denote valid sequences with curly braces) of $q_1\ and \ q_2\ are \ \{0\},1,\{2,3,4\},5,\{6\},7,\{8,9,10\},11,\{12\}\dots$ As you can see from example, the available values for $l$ are finite, because they repeat themselves after $6,12\dots$ The Longest sequence is $\{1,2,3\}, \{8,9,10\}\dots$ and I expect the answer of $l$ to be $3$
If the $q_i$ are coprime the pattern will repeat after the product of all the $q_i$, so if you are doing brute force you can stop there. In your example with $2,3$ you could stop after $6$ and know you have the longest consecutive sequence.
You can do better, but it is hard to describe as an algorithm. Let's say we have the $q_i 2,3,5,13$. This will not repeat until $390$. We can make a row $$2\_2\_2\_2\_2\_2\_2\_2\_2\_2\_2\_2\_2\_2\_2\_2\_2$$ which shows that $2$ divides every other number but the alternate ones are not accounted for yet. Now fill in a blank with $3$ and note that it will repeat every $6$. There is another multiple of $3$ in the middle, but that number is already a multiple of $2$. That gives $$232\_2\_232\_2\_232\_2\_232\_2\_232\_2\_232$$ Now we note that the $5$s will come $10$ apart and there are always at least two blanks between them. Putting a $5$ in the blank right after a $3$ makes the next occurence of $5$ land on a $3$, so we wait one. We can only fill one of those with a $13$, so it becomes $$232(13)25232\_2\_23252\_232\_2\_232\_2\_232$$ which shows we can get nine numbers in order with this set. The Chinese remainder theorem guarantees this pattern exists somewhere because the first number is the solution to $$n\equiv 0 \pmod 2\\n\equiv 2 \pmod 3\\n\equiv 0 \pmod 5\\n \equiv 10 \pmod {13}$$