I had a simpler question before such that I could even answer it myself. For the next step I seem again to be too dense today. (Remark several days later: it's not only being dense... I still don't find the first step for the solution)
Recall: I discuss q,r as residues to a modulus of $\small 2^B$ for natural parameters B. I assume an odd $\small r (\gt 0) $ as given and q as $\small q= \frac1r \pmod {2^B}$. I understand now well, that $\small {qr-1\over 2^B} +1 \le \min(q,r) $
But I observe more: I find in some experiments using Pari/GP, that the equality occurs exactly iff either r or q is a divisor (or both are divisors) of $\small 2^B-1 $.
How can I show this with a proof?
Examples.
We use $\small B=8, 2^B=256$
First we try $\small r=15$. Then $\small 1/15 \equiv 239 \pmod{256} \to q=239$
Also r is a divisor of $\small 256-1 $ . Then$\qquad \small {15\cdot239-1\over256}+1 =15 = \min(15,239)$
Next we try $\small r=13$. Then $\small 1/13 \equiv 197 \pmod{256} \to q=197$
Now r is not a divisor of $\small 256-1 $ Then$\qquad \small {13\cdot197-1\over256}+1 =10 \lt \min(13,197) $
Here is some Pari/GP-code to see what I mean
B=9 \\ chose some exponent B Test(B) \\ check display {Test(B) = local(M,M1,r,q,t,rhs,isdiv); M = 2^B ; M1 = 2^B-1 ; for(k=1,M/2, r=2*k-1; \\ test all odd residues up to 2^B-1 q=1 / r % M ; \\ q is the multiplicative inverse (mod 2^B) t = (r*q-1)/M +1 ; rhs = min(r,q); isdiv = ((M1 % r ) * (M1 % q)) == 0 ; \\ =1 if either q or r is divisor of 2^B-1 print([r,q,t, rhs, t == rhs, isdiv]); )}