1/a + 1/b + 1/c invariant....

As in post #2, notice that $\frac{ab+bc+ca}{abc}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$. Thus, if the set of numbers on the board is $\mathcal{S}$, $\sum_{n\in \mathcal{S}}\frac{1}{n}$ remains invariant. Notice that at the beginning this quantity is $$\sum_{i=2}^{2n+1}\frac{1}{i^3-i}=\sum_{i=2}^{2n+1}\frac{1}{2}\left(\frac{1}{i+1}-\frac{1}{i-1}\right)-\frac{1}{i}=\frac{2n^2+3n}{8n^2+12n+4}$$If the numbers at the end are $a$ and $b$, we hence have $\frac{a+b}{ab}=\frac{2n^2+3n}{8n^2+12n+4}=\frac{1}{4}\frac{2n^2+3n}{2n^2+3n+1}$. Notice the last fraction is irreducible by Euclid, so $ab\ge 2n^2+3n+1$. Thus $a+b\ge 2\sqrt{ab}\ge 2\sqrt{2n^2+3n+1}$ by AM-GM. This solves the problem for $n\ge 5$. It suffices to handle $n=2,3,4$ which is easy.

Sniped, Call $a_i = i^3-i$ for all $i = 2,3,\ldots, 2n+1$. $$X = \sum_{i=2}^{2n+1}\frac{1}{a_i}$$is an invariant. More precise, $$X = \frac{2n^2+3n}{8n^2+12n+4} < \frac{1}{4}$$. Suppose $b$ and $c$ are the last two numbers on the board. Suppose $b+c \le 16$. Cauchy gives: $$(1+1)^2 \le (b+c)(\frac{1}{b} + \frac{1}{c}) = (b+c) \cdot X \le 16X \iff X \ge \frac{1}{4}$$, which is a contradiction. Therefore, the sum of the last two numbers is greater than $16$.