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This is too long for a comment but far from an answer.

Edited version: Just to be clear, the question is: can we find integers $n$ and $s$ and arithmetic progressions $r_{n+i} \mod p_{n+i}$ where i ranges from 1 to the largest $j$ such that $p_{n+j} \lt P_n$ such that the arithmetic progressions cover the interval $[s,s+P_n-1]$? I would guess that it IS possible (but I could be wrong). However it would definitely likely require a huge number of progressions. A better question might be given a certain set of primes (or pairwise relatively prime moduli), what is the longest interval you can cover?

There is no harm in assuming that $s=0$ since we can instead look at the progressions $r_{n+i}-s \mod p_{n+i}$.

If it can be done at all then the number of ways to do it (provided that we do get to pick $s$ to be what we want) is simply the product of all the primes in the range: Whatever residue classes you choose will create a pattern of covered and uncovered integers which is periodic but with an extremely long period. Picking different residue classes creates the same pattern, just shifted.

A question which makes some sense is: Bound the number of ways to choose residue classes and cover $[0,P_n-1]$ . Even that would be huge. One would have some carefully chosen progressions for the smaller primes and also an enormous number of one member residue classes filling in the holes. Those one member progressions could be shuffled around at will.

We would certainly need to have $\sum_1^j\lceil\frac{P_n}{p_{n+i}}\rceil \gt P_n$ but this is on the overly optimistic chance that we could have all the progressions distinct (At least for $p_{n+i} \lt \sqrt{P_n}$ the progressions will not be completely disjoint) and that every one could be positioned to get in the maximum number of occurrences.

A condition which does not make the second assumption is $\sum_1^j\frac{1}{p_{n+i}} \gt 1$ so $\sum_1^{n+j}\frac{1}{p_{k}} \gt 1+\sum_1^{n}\frac{1}{p_{k}}$. By my calculations using this estimate $$\sum_{p \lt x}\frac1p \ge \ln \ln (x+1) - \ln\frac{\pi^2}6$$ that condition requires that $p_n$ is at least $23$ meaning that $P_n$ is greater than $2.23\cdot 10^8$ and $j>1.2 \cdot 10^7$. This makes the greedy strategy unattractive (Start with $s=0$ (why not?) and pick residue classes repeatedly to take care of the smallest uncovered integer). At any rate, that argument ignores the fact that at least for $p_{n+i} \lt \sqrt{P_n}$ the progressions will not be completely disjoint. Perhaps $\prod_1^j(1-\frac{1}{p_{n+i}})\ll \frac{1}{P_n}$ is a better condition. Taking that into account is a more delicate endeavor than I care to undertake at the moment.

I think that if (for a given $n$ and $s$) some choice of residue classes works, then any choice of classes works for that same $n$ with an appropriate change in $s$ AND any $s$ works with an appropriate change in the residue classes (using the CRT)

later thoughts In a comment below Gerghard shows how to cover the interval $[0,73]$ with the $42$ primes from $11$ to $199$. Since those primes have reciprocals adding to under 0.8 I should adjust a claim above. Even though $\frac{69}{17}$ is barely over $4$, the congruence class $0 \mod 17$ has $5$ members. So I should say that to cover an interval of length $L$ requires that $\sum_1^{n+j}\lceil \frac{L}{p_{k}}\rceil \ge L.$ That sum is 77 for the example.

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This is too long for a comment but far from an answer. Just to be clear, the question is: can we find integers $n$ and $s$ and arithmetic progressions $r_{n+i} \mod p_{n+i}$ where i ranges from 1 to the largest $j$ such that $p_{n+j} \lt P_n$ such that the arithmetic progressions cover the interval $[s,s+P_n-1]$? I would guess that it IS possible (but I could be wrong). However it would definitely require a huge number of progressions:

We would certainly need to have $\sum_1^j\frac{1}{p_{n+i}} \gt 1$ so $\sum_1^{n+j}\frac{1}{p_{k}} \gt 1+\sum_1^{n}\frac{1}{p_{k}}$. By my calculations using this estimate $$\sum_{p \lt x}\frac1p \ge \ln \ln (x+1) - \ln\frac{\pi^2}6$$ that condition requires that $p_n$ is at least $23$ meaning that $P_n$ is greater than $2.23\cdot 10^8$ and $j>1.2 \cdot 10^7$. This makes the greedy strategy unattractive (Start with $s=0$ (why not?) and pick residue classes repeatedly to take care of the smallest uncovered integer). At any rate, that argument ignores the fact that at least for $p_{n+i} \lt \sqrt{P_n}$ the progressions will not be completely disjoint. Perhaps $\prod_1^j(1-\frac{1}{p_{n+i}})\ll \frac{1}{P_n}$ is a better condition. Taking that into account is a more delicate endeavor than I care to undertake at the moment.

I think that if (for a given $n$ and $s$) some choice of residue classes works, then any choice of classes works for that same $n$ with an appropriate change in $s$ AND any $s$ works with an appropriate change in the residue classes (using the CRT)

later thoughts In a comment below I show Gerghard shows how to cover the interval $[0,69]$ [0,73]$with the$42$primes from$11$to$199$. Since those primes have reciprocals adding to about 0.5 under 0.8 I should adjust a claim above. Even though$\frac{69}{17}$is barely over$4$, the congruence class$0 \mod 17$has$5$members. So I should say that to cover an interval of length$L$requires that$\sum_1^{n+j}\frac{1}{p_{k}}>1-\frac{j}{L}.$\sum_1^{n+j}\lceil \frac{L}{p_{k}}\rceil \ge L.$ That sum is 77 for the example.

2 added 425 characters in body

This is too long for a comment but far from an answer. Just to be clear, the question is: can we find integers $n$ and $s$ and arithmetic progressions $r_{n+i} \mod p_{n+i}$ where i ranges from 1 to the largest $j$ such that $p_{n+j} \lt P_n$ such that the arithmetic progressions cover the interval $[s,s+P_n-1]$? I would guess that it IS possible (but I could be wrong). However it would definitely require a huge number of progressions:

We would certainly need to have $\sum_1^j\frac{1}{p_{n+i}} \gt 1$ so $\sum_1^{n+j}\frac{1}{p_{k}} \gt 1+\sum_1^{n}\frac{1}{p_{k}}$. By my calculations using this estimate $$\sum_{p \lt x}\frac1p \ge \ln \ln (x+1) - \ln\frac{\pi^2}6$$ that condition requires that $p_n$ is at least $23$ meaning that $P_n$ is greater than $2.23\cdot 10^8$ and $j>1.2 \cdot 10^7$. This makes the greedy strategy unattractive (Start with $s=0$ (why not?) and pick residue classes repeatedly to take care of the smallest uncovered integer). At any rate, that argument ignores the fact that at least for $p_{n+i} \lt \sqrt{P_n}$ the progressions will not be completely disjoint. Perhaps $\prod_1^j(1-\frac{1}{p_{n+i}})\ll \frac{1}{P_n}$ is a better condition. Taking that into account is a more delicate endeavor than I care to undertake at the moment.

I think that if (for a given $n$ and $s$) some choice of residue classes works, then any choice of classes works for that same $n$ with an appropriate change in $s$ AND any $s$ works with an appropriate change in the residue classes (using the CRT)

later thoughts In a comment below I show how to cover the interval $[0,69]$ with the $42$ primes from $11$ to $199$. Since those primes have reciprocals adding to about 0.5 I should adjust a claim above. Even though $\frac{69}{17}$ is barely over $4$, the congruence class $0 \mod 17$ has $5$ members. So I should say that to cover an interval of length $L$ requires that $\sum_1^{n+j}\frac{1}{p_{k}}>1-\frac{j}{L}.$

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