jmc

Since the numbers $r+\frac{19}{100},$ $r+\frac{20}{100},$ $r+\frac{21}{100},$ $\dots ,$ $r+\frac{91}{100}$ 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, $$\lfloor r+\frac{19}{100}\rfloor =\lfloor r+\frac{20}{100}\rfloor =\cdots =\lfloor r+\frac{56}{100}\rfloor =7$$and $$\lfloor r+\frac{57}{100}\rfloor =\lfloor r+\frac{58}{100}\rfloor =\cdots =\lfloor r+\frac{91}{100}\rfloor =8.$$In particular, $r+\frac{56}{100}<8\le r+\frac{57}{100},$ so $7.43\le r<\mathrm{7.44.}$ Thus, $743\le 100r<744,$ so the answer is $$\lfloor 100r\rfloor =\boxed{743}.$$