The nineteenth question: Find the smallest real number $\lambda$ such that $\sum_{i=1}^{100}\left(a_{i}-a_{i+1}\right)^{2} \leq \lambda\left(100-\left(\sum_{i=1}^{100} a_{i}\right)\right)$ holds for any real numbers $a_{1} 、 a_{2} 、 \ldots 、 a_{100}$ satisfying $\sum_{i=1}^{100} a_{i}^{2}=100$. Here $a_{101}=a_{1}$.
Nineteenth question,
Solution: On the one hand, take \(a_{1}=a_{3}=\ldots=a_{99}=\sqrt{1+\varepsilon}, a_{2}=a_{4}=\ldots=a_{100}=\sqrt{1-\varepsilon}\), where \(\varepsilon \in[-1,1]\), then
\[
\begin{array}{l}
100(\sqrt{1+\varepsilon}-\sqrt{1-\varepsilon})^{2} \leq 50 \lambda(2-(\sqrt{1+\varepsilon}+\sqrt{1-\varepsilon})) \\
\Rightarrow \lambda \geq \frac{2(\sqrt{1+\varepsilon}-\sqrt{1-\varepsilon})^{2}}{2-(\sqrt{1+\varepsilon}+\sqrt{1-\varepsilon})}
\end{array}
\]

Notice that \(\sqrt{1+\varepsilon}+\sqrt{1-\varepsilon} \leq 2\), and \((\sqrt{1+\varepsilon}+\sqrt{1-\varepsilon})^{2}+(\sqrt{1+\varepsilon}-\sqrt{1-\varepsilon})^{2}=4\). Let \(\sqrt{1+\varepsilon}+\sqrt{1-\varepsilon}=2-\omega\), then \((\sqrt{1+\varepsilon}-\sqrt{1-\varepsilon})^{2}=4-(2-\omega)^{2}\), and when \(\varepsilon \rightarrow 0\), \(\omega \rightarrow 0^{+}\). Thus, we know:
\[
\lambda \geq \lim _{\varepsilon \rightarrow 0} \frac{2(\sqrt{1+\varepsilon}-\sqrt{1-\varepsilon})^{2}}{2-(\sqrt{1+\varepsilon}+\sqrt{1-\varepsilon})}=\lim _{\omega \rightarrow 0^{+}} \frac{2\left(4-(2-\omega)^{2}\right)}{\omega}=8
\]

Next, we prove:
\[
\sum_{i=1}^{100}\left(a_{i}-a_{i+1}\right)^{2} \leq 8\left(100-\left(\sum_{i=1}^{100} a_{i}\right)\right)
\]

In fact,
\[
\begin{array}{l}
8\left(100-\left(\sum_{i=1}^{100} a_{i}\right)\right)=8\left(\sum_{i=1}^{100} a_{i}^{2}-\frac{\left(\sum_{i=1}^{100} a_{i}\right) \sqrt{\sum_{i=1}^{100} a_{i}^{2}}}{10}\right) \\
=\frac{2}{25}\left(100 \sum_{i=1}^{100} a_{i}^{2}-\left(\sum_{i=1}^{100} a_{i}\right) \sqrt{100 \cdot \sum_{i=1}^{100} a_{i}^{2}}\right) \\
=\frac{2}{25} \cdot \frac{\sqrt{100 \cdot \sum_{i=1}^{100} a_{1}^{2}} \cdot\left(100 \sum_{i=1}^{100} a_{i}^{2}-\left(\sum_{i=1}^{100} a_{i}\right)^{2}\right)}{\sqrt{100 \sum_{i=1}^{100} a_{i}^{2}}+\left(\sum_{i=1}^{100} a_{i}\right)} \\
\geq \frac{100 \sum_{i=1}^{100} a_{i}^{2}-\left(\sum_{i=1}^{100} a_{i}\right)^{2}}{25}
\end{array}
\]

Therefore, we only need to prove:
\[
\sum_{i=1}^{100}\left(a_{i}-a_{i+1}\right)^{2} \leq \frac{100 \sum_{i=1}^{100} a_{i}^{2}-\left(\sum_{i=1}^{100} a_{i}\right)^{2}}{25}
\]

Notice that \(100 \sum_{i=1}^{100} a_{i}^{2}-\left(\sum_{i=1}^{100} a_{i}\right)^{2}=\sum_{1 \leq i<j \leq 100}\left(a_{i}-a_{j}\right)^{2}\), by translation invariance, we can assume \(\sum_{i=1}^{100} a_{i}=0\), then we only need to prove:
\[
\sum_{i=1}^{100}\left(a_{i}-a_{i+1}\right)^{2} \leq 4 \sum_{i=1}^{100} a_{i}^{2}
\]

In fact,
\[
\sum_{i=1}^{100}\left(a_{i}-a_{i+1}\right)^{2} \leq 4\left(\sum_{i=1}^{100} a_{i}^{2}\right) \Leftrightarrow \sum_{i=1}^{100}\left(a_{i}+a_{i+1}\right)^{2} \geq 0
\]

The above inequality is obviously true.
In conclusion, the minimum value of \(\lambda\) is 8.