Solution:
We have $a_{1}+a_{2}+\cdots+a_{100} \leq 200$, so
$$\begin{aligned} a_{1}^{2}+a_{2}^{2}+\cdots+a_{100}^{2} & \leq (100-a_{2})^{2}+a_{2}^{2}+a_{3}^{2}+\cdots+a_{100}^{2} \\ & = 100^{2}-200 a_{2}+2 a_{2}^{2}+a_{3}^{2}+\cdots+a_{100}^{2} \\ & \leq 100^{2}-(a_{1}+a_{2}+\cdots+a_{100}) a_{2}+2 a_{2}^{2}+a_{3}^{2}+\cdots+a_{100}^{2} \\ & = 100^{2}+(a_{2}^{2}-a_{1} a_{2})+(a_{3}^{2}-a_{3} a_{2})+(a_{4}^{2}-a_{4} a_{2})+\cdots+(a_{100}^{2}-a_{100} a_{2}) \\ & = 100^{2}+(a_{2}-a_{1}) a_{2}+(a_{3}-a_{2}) a_{3}+(a_{4}-a_{2}) a_{4}+\cdots+(a_{100}-a_{2}) a_{100} \end{aligned}$$
Since $a_{1} \geq a_{2} \geq \cdots \geq a_{100} \geq 0$, none of the terms $(a_{i}-a_{j}) a_{i}$ is positive. Thus $a_{1}^{2}+a_{2}^{2}+\cdots+a_{100}^{2} \leq 10,000$ with equality holding if and only if
$$a_{1}=100-a_{2} \quad \text{ and } \quad a_{1}+a_{2}+\cdots+a_{100}=200$$
and each of the products
$$(a_{2}-a_{1}) a_{2}, \quad (a_{3}-a_{2}) a_{3}, \quad (a_{4}-a_{2}) a_{4}, \cdots, (a_{100}-a_{2}) a_{100}$$
equals zero. Since $a_{1} \geq a_{2} \geq a_{3} \geq \cdots \geq a_{100} \geq 0$, the last condition holds if and only if for some $i \geq 1$ we have $a_{1}=a_{2}=\cdots=a_{i}$ and $a_{i+1}=\cdots=a_{100}=0$. If $i=1$, then we get the solution $100,0,0,\ldots,0$. If $i \geq 2$, then from $a_{1}+a_{2}=100$, we get that $i=4$ and the second optimal solution $50,50,50,50,0,0, \ldots, 0$.