If you pick a natural number n, for example 3, and you take the string from 1,..,n and shift it via 2,...,n+1; 3, ... n+2 etc, then it looks like you will first find some strings where you can do the following:
-Take the numbers of 1,...,n.
-If i is a divisor of a j taken from your new string k,..., n+k, you can (if you want) 'couple' i to that particular j.
-You might be able to UNIQUELY couple all 1,...,n to a certain k,..., n+k (call it a bijection if you want).
Example: 1,2,3 can be 'coupled' with 3,4,5 because (3,3), (2,4) and (1,5) satisfy.
But if you keep shifting the string, it look like their will always be a string where you can't do this. In the case of n = 3: (5,6,7) won't work for the simple reason that 5 and 7 are prime.
It looks like there is some kind of a pattern in, for a given n, the first string in which you cannot do this. I've tried quite a lot of them (some might be wrong):
1: infinity, since 1 divides all numbers 2: infinity, since all two consecutive numbers contain one even number 3: (567) 4: (4-7) 5: (7-11) 6: (6-11) 7: (7-13) 8: (6-13) 9: (5-13) 10:(4-13) 11: (7-17) 12: (6-17) 13: (7-19) 14: (6-19) 15: (5-19) 16: (4-19) 17: (6-22) 18: (6-23) 19: (7-26) 20: (6-26) 21: (5-25) 22: (4-25) 23: (9-31) 24: (8-31) 25: (7-31) 26: (6-31) 27: (5-31) 28: (4-31)
Is there a reason why it has this specific shape i.e. why the last digit, which seemed to be prime always, suddenly become composite? This will, for larger n, repeat (e.g. 31), but I haven't found a number with more than 2 different primes in the prime-factorisation. Will they occur for even larger n? Notice that the last digit only seemed to change when the chosen natural number is prime.
Writing a programme to calculate the first string for a given n crashed about n=10.
An upper limit to the top of the range is the second prime above $n$, call it $p$. As you say, this is because one prime can be associated with $1$ but the second prime cannot be associated with anything. The lower limit of the range would then be $p-n+1$.
The first place this bound is not achieved is $17$ where the upper limit is $22$ instead of $23$. If we try to make the assignment you need $1$ for $19$, $11$ for $11$, so have to use $2$ for $22$. We have to use $7$ for $7$, so $3$ for $21$, and don't have anything for $6$. This sort of chain will work whenever $6$ plus twice a prime is greater than $n$ and less than the second prime above $n$.
An interesting one to try is $31$ because there are so few primes close above it. I believe $(8,38)$ fails because you can't use both $5$ and $7$ for $35$