jmc

Let $$f(n)=\lceil \frac{99n}{100}\rceil -\lfloor \frac{100n}{101}\rfloor .$$Note that $$\begin{array}{rl}f(n+10100)& =\lceil \frac{99(n+10100)}{100}\rceil -\lfloor \frac{100(n+10100)}{101}\rfloor \\ & =\lceil \frac{99n}{100}+101\rceil -\lfloor \frac{100n}{101}+100\rfloor \\ & =\lceil \frac{99n}{100}\rceil +101-\lfloor \frac{100n}{101}\rfloor -100\\ & =\lceil \frac{99n}{100}\rceil -\lfloor \frac{100n}{101}\rfloor +1\\ & =f(n)+1.\end{array}$$This implies that for each residue class $r$ modulo 10100, there is a unique integer $n$ such that $f(n)=1$ and $n\equiv r(\mathrm{mod}10100).$ Thus, the answer is $\boxed{10100}.$