Problem 1624

32. Let $f(x)=\left(x-a_{1}\right)^{2}+\left(x-a_{2}\right)^{2}+\cdots+\left(x-a_{n}\right)^{2}$

Then $\square$
$$f(x)=n\left(x-a_{n}\right)^{2}+f\left(b_{n}\right)$$

Now we prove $C_{n} \leqslant D_{n} \leqslant 2 C_{n}$ by mathematical induction.
When $n=1$, $C_{1} \leqslant D_{1}$, so $C_{1} \leqslant D_{1} \leqslant 2 C_{1}$.
Assume that when $n$, the inequality holds. When adding a number $a_{n+1}$ to $a_{1}, a_{2}, \cdots, a_{n}$, $C_{n}$ increases by $\left(a_{n+1}-b_{n+1}\right)^{2}$, and $D_{n}$ increases by $\left(a_{n+1}-b_{n+1}\right)^{2}+f\left(b_{n+1}\right)-f\left(b_{n}\right)$.

In equation (1), let $x=b_{n+1}$, we get
$$0 \leqslant f\left(b_{n+1}\right)-f\left(b_{n}\right)=n\left(b_{n+1}-b_{n}\right)^{2}=\frac{1}{n}\left(a_{n+1}-b_{n+1}\right)^{2} \leqslant\left(a_{n+1}-b_{n+1}\right)^{2}$$

Thus, the value by which $D_{n}$ increases, $\left(a_{n+1}-b_{n+1}\right)^{2}+f\left(b_{n+1}\right)-f\left(b_{n}\right)$, lies between $\left(a_{n+1}-b_{n+1}\right)^{2}$ and $2\left(a_{n+1}-b_{n+1}\right)^{2}$. Therefore, for $n+1$, we also have $C_{n+1} \leqslant D_{n+1} \leqslant 2 C_{n+1}$.

Hence, for all positive integers $n$, $C_{n} \leqslant D_{n} \leqslant 2 C_{n}$.