Solution:

The answer is (D). Let us denote by $\mathcal{A}$ the set of integers $n$ satisfying the required property. Let us first observe that: $\sqrt{101}-\sqrt{81}>\sqrt{100}-\sqrt{81}=1$, that is $81 \notin \mathcal{A}$.

If $n$ is a positive integer, the difference
$$\sqrt{n+1}-\sqrt{n}=\frac{1}{\sqrt{n+1}+\sqrt{n}}$$
between the square roots of two consecutive integers decreases as $n$ increases. Consequently $\sqrt{82}-\sqrt{81}>\sqrt{101}-\sqrt{100}$, that is $\sqrt{101}-\sqrt{82}<\sqrt{100}-\sqrt{81}=1$, and therefore $82 \in \mathcal{A}$; In the same way one shows that $122 \in \mathcal{A}$, in fact $\sqrt{122}-\sqrt{121}<\sqrt{101}-\sqrt{100}$, from which $\sqrt{122}-\sqrt{101}<\sqrt{121}-\sqrt{100}=1$. To show that $123 \notin \mathcal{A}$, one can do the brute calculation $(\sqrt{123}-\sqrt{101})^{2}=123+101-2 \cdot \sqrt{123 \cdot 101}$, that is one must show that $224>2 \cdot \sqrt{12423}$ that is that $\sqrt{12423}<112$; this is true because $112^{2}=12544>12423$.

Ultimately, $\mathcal{A}=\{n \in \mathbb{N}: 82 \leq n \leq 122\}$ and it is made up of $122-82+1=41$ elements.