Using the knowledge from sections $\S 1$ and $\S 2$ of Chapter 1, we can prove (i). The proof of (i) relies on the fundamental properties of the divisors of a positive integer $n$: If $d_{1}, d_{2}, \cdots, d_{k}$ are all the (positive) divisors of $n$, then $n / d_{1}, n / d_{2}, \cdots, n / d_{k}$ are also all the (positive) divisors of $n$, and when $1=d_{1}<d_{2}<\cdots<d_{k}=n$, we have $d_{j}=n / d_{k-j+1}$ (see Theorem 2 in section $\S 2$ of Chapter 1). Using this property and the given conditions, we get
$$\begin{aligned} n^{2} / d_{2}= & n^{2} /\left(d_{1} d_{2}\right) \leqslant D=d_{1} d_{2}+d_{2} d_{3}+\cdots+d_{k-1} d_{k} \\ = & n^{2}\left\{1 /\left(d_{k-1} d_{k}\right)+1 /\left(d_{k-2} d_{k-1}\right)+\cdots+1 /\left(d_{1} d_{2}\right)\right\} \\ \leqslant & n^{2}\left\{\left(1 / d_{k-1}-1 / d_{k}\right)+\left(1 / d_{k-2}-1 / d_{k-1}\right)\right. \\ & \left.+\cdots+\left(1 / d_{1}-1 / d_{2}\right)\right\} \\ = & n^{2}\left(1 / d_{1}-1 / d_{k}\right)=n^{2}(1-1 / n)=n^{2}-n . \end{aligned}$$

This proves (i). The leftmost inequality in the above expression holds with equality if and only if $k=2$, i.e., the only positive divisors of $n$ are 1 and $n$ itself, which means $n$ is a prime number (see Definition 2 in section 2.2 of Chapter 1). Additionally, the rightmost inequality holds with equality if and only if $n=2$ (why).