Since the numbers $ r + \tfrac{19}{100}, $ $ r+\tfrac{20}{100}, $ $ r+\tfrac{21}{100}, $ $ \ldots, $ $ r+\tfrac{91}{100} $ are all (strictly) within $ 1 $ of each other, the first few terms on the left-hand side must all equal some integer $ n, $ and all the other terms (if any) must equal $ n+1 $. There are $ 91 - 19 + 1 = 73 $ terms on the left-hand side. We have $ 546 = 7 \cdot 73 + 35, $ which shows that $ n=7, $ and that $ 35 $ of the terms equal $ 8, $ while the first $ 73 - 35 = 38 $ terms equal $ 7 $. Thus, \[\left\lfloor r + \frac{19}{100} \right\rfloor = \left\lfloor r + \frac{20}{100} \right\rfloor = \dots = \left\lfloor r + \frac{56}{100} \right\rfloor = 7\]and \[\left\lfloor r + \frac{57}{100} \right\rfloor = \left\lfloor r + \frac{58}{100} \right\rfloor = \dots = \left\lfloor r + \frac{91}{100} \right\rfloor = 8.\]In particular, $ r + \tfrac{56}{100} < 8 \le r + \tfrac{57}{100}, $ so $ 7.43 \le r < 7.44 $. Thus, $ 743 \le 100r < 744, $ so the answer is \[\lfloor 100r \rfloor = \boxed{743}.\]