Example 1 If the real number $r$ satisfies $\left[r+\frac{19}{100}\right]+\left[r+\frac{20}{100}\right]+\cdots+\left[r+\frac{91}{100}\right]=$ 546, find $[100 r]$.
Solution: The left side of the equation has 73 terms, and since $\frac{19}{100}, \frac{20}{100}, \cdots, \frac{91}{100}$ are all less than 1, each term is either $[r]$ or $[r+1]$. However, note that $73 \times 7 < 546 < 73 \times 8$, so $[r]$ must be 7.

Let $\left[r+\frac{k}{100}\right]=7(19 \leqslant k \leqslant m),\left[r+\frac{p}{100}\right]=8(m+1 \leqslant p \leqslant 91)$, then $7(m-18)+8(91-m)=546$, solving for $m$ gives $m=56$, hence $\left[r+\frac{56}{100}\right]=7$, which means $6.44 \leqslant r < 7.44$; $\left[r+\frac{57}{100}\right]=8$, which means $7.43 \leqslant r < 8.43$. Therefore, $7.43 \leqslant r < 7.44$. So $[100 r]=743$.
$\left.\frac{n-1}{n}\right]$, then $f\left(x+\frac{1}{n}\right)=[n x]+1-\left[x+\frac{1}{n}\right]-\left[x+\frac{2}{n}\right]-\cdots-[x+$ $\left.\frac{n-1}{n}\right]-[x]-1=f(x)$, which means $f(x)$ is a periodic function with period $\frac{1}{n}$.
Therefore, we only need to prove: when $x \in\left[0, \frac{1}{n}\right]$, $f(x)=0$.
By the definition of the floor function, it is clear that $f(x)=0, x \in\left[0, \frac{1}{n}\right]$, so by the periodicity of $f(x)$, we can conclude that for all $x \in \mathrm{R}$, $f(x)=0$, i.e., the equation holds.