Since $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{ab+bc+ca}{abc} = \left(\frac{abc}{ab+bc+ca}\right)^{-1}$, for any given positive numbers $a$, $b$, $c$, it follows that after any operation, the sum of the inverses of the numbers left on the board is equal to the sum of the inverses of the numbers on the board before the operation. Therefore, the sum of the inverses of the last two numbers left on the board is equal to the sum $S$ of the inverses of the initial numbers.

Notice that $\frac{1}{k^3-k} = \frac{1}{k(k-1)(k+1)} = \frac{1}{2}\left(\frac{1}{(k-1)k} - \frac{1}{k(k+1)}\right)$, for any $k > 1$, so:
$$\begin{align*} S &= \frac{1}{2^3 - 2} + \frac{1}{3^3 - 3} + \dots + \frac{1}{(2n+1)^3 - (2n+1)} \\ &= \frac{1}{2} \left( \frac{1}{1 \cdot 2} - \frac{1}{2 \cdot 3} + \frac{1}{2 \cdot 3} - \frac{1}{3 \cdot 4} + \dots + \frac{1}{2n(2n+1)} - \frac{1}{(2n+1)(2n+2)} \right) \\ &= \frac{1}{2} \left( \frac{1}{2} - \frac{1}{(2n+1)(2n+2)} \right) = \frac{2n^2 + 3n}{8n^2 + 12n + 4} \end{align*}$$
If $x$ and $y$ are the last two numbers left on the board, then $\frac{1}{x} + \frac{1}{y} = \frac{x+y}{xy} = \frac{2n^2+3n}{8n^2+12n+4}$. Since $\frac{4}{x+y} \le \frac{x+y}{xy}$, for each $x$, $y > 0$, it follows that $\frac{4}{x+y} \le \frac{2n^2+3n}{8n^2+12n+4} < \frac{1}{4}$, so $x+y > 16$.