Question 210, Find the number of integers $n$ that satisfy $1+\left\lfloor\frac{100 n}{101}\right\rfloor=\left\lceil\frac{99 n}{100}\right\rceil$, where $\lfloor x]$ denotes the greatest integer not greater than $x$, and $[\mathrm{x}]$ denotes the smallest integer not less than $\mathrm{x}$.
Question 210, Solution: Let $f(n)=\left\lceil\frac{99 n}{100}\right\rceil-\left\lfloor\frac{100 n}{101}\right\rfloor$, we need to find the number of integers that satisfy $f(n)=1$. Notice that
$$
\begin{array}{l}
f(n+10100)=\left\lceil\frac{99(n+10100)}{100}\right\rceil-\left\lfloor\frac{100(n+10100)}{101}\right\rfloor \\
=\left(\left\lceil\frac{99 n}{100}\right\rceil+99 \times 101\right)-\left(\left\lfloor\frac{100 n}{101}\right\rfloor+100 \times 100\right)=f(n)-1
\end{array}
$$

Therefore, for each residue class modulo 10100, which is $r (r=0, 1, 2, \ldots, 10099)$, there is exactly one integer $n$ such that $f(n)=1$. Thus, the number of integers $n$ that satisfy the condition is 10100.