Let's find the fraction with the smallest denominator $p / q$; $p, q$ are positive integers; for which

$$
\frac{99}{100}<\frac{p}{q}<\frac{100}{101}
$$
We will show that if a fraction satisfies

$$
\frac{99}{100}<\frac{p}{q}<\frac{100}{101}
$$

then the denominator of the fraction is at least 100+101, the sum of the denominators. To do this, we multiply the (1) inequalities by the product of the denominators:

$$
99 \cdot 101 q<100 \cdot 101 p<100^{2} q
$$

The first two numbers are multiples of 101, so their difference is at least 101; similarly, the difference between the second two numbers is at least 100. Thus,

$$
100^{2} q-99 \cdot 101 q \geqq 100+101=201
$$

The left side of the (3) inequality is exactly

$$
\left(100^{2}-99 \cdot 101\right) q=q
$$

and thus we obtain the statement to be proven.

We also need to show that there exists a fraction with a denominator of 201 that satisfies (1). Since

$$
\frac{99}{100}=\frac{198}{200} \text { and } \frac{100}{101}=\frac{200}{202} \text { and } \frac{198}{200}<\frac{199}{201}<\frac{200}{202}
$$

it follows that (1) is satisfied with $p=199$ if $q=201$,

Thus, the sought fraction is $\frac{199}{201}$. (This is the only fraction with a denominator of 201 that satisfies the conditions, since the difference between two such fractions is at least $\frac{1}{201}$, and $\frac{100}{101}-\frac{99}{100}=\frac{1}{100.101}<\frac{1}{201}$.)