Let's find the positive integer \( n \) for which

\[
\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}+\sqrt{n+1}}=100
\]
Let's rationalize the fractions, and expand each by the "conjugate" of their denominators:

So

$$
\begin{gathered}
\frac{1}{\sqrt{k}+\sqrt{k+1}}=\frac{\sqrt{k+1}-\sqrt{k}}{(\sqrt{k+1}-\sqrt{k})(\sqrt{k+1}+\sqrt{k})}=\sqrt{k+1}-\sqrt{k} \\
\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+\cdots+\sqrt{n+1}-\sqrt{n}=100
\end{gathered}
$$

From here, $\sqrt{n+1}-1=100$, and $n=10200$.

Paulin Dániel (Fazekas M. Főv. Gyak. Gimn., 8th grade)