How can this be proven?
If we list the divisors of a natural number $N$, and pick the two in the middle, and then multiply them, we get $N$.
If $N$ has an even amount of divisors, then we pick the two in the middle, and if it has an odd amount of divisors, we pick the middle one repeated twice. $$\operatorname{Divisors}(9) = (1, 3, 9) \longrightarrow 3 \cdot 3 = 9$$ $$\operatorname{Divisors}(24) = (1, 2, 3, 4, 6, 8, 12, 24) \longrightarrow 4\cdot6 = 24 $$ And so on...
If p divides n then q = n/p also divides n. So each divisor p can be paired with divisor n/p. If you arrange the divisors, $d_i$ in order of size you get.
$1 = d_0 = \frac n {d_n} < d_0 =\frac n{d_{n-1}} < d_2 = \frac n{d_{n-2}} < ... < d_{n-2} = \frac n{d_2} < d_{n-1} = \frac n{d_1} < d_n = \frac n{d_0}= \frac n 1 = n$
(Because $d_i < d_j$ iff $d_{n - i} = \frac {n}{d_i} > d_{n - j} = \frac n {d_j}$.)
In all cases $d_i * d_{n-i} = n$.