Let $a_0, b_0, c_0$ be the initial numbers on the blackboard, and $a_i, b_i, c_i$ be the numbers written on the blackboard after the $i$-th move; let $S(a_0, b_0, c_0)$ and $S(a_i, b_i, c_i)$ be the sums of these numbers, respectively. Then after the $i+1$-th move we obtain the following numbers
$$a_{i+1} = \frac{b_i^2 + c_i^2}{a_i}, \quad b_{i+1} = \frac{c_i^2 + a_i^2}{b_i}, \quad c_{i+1} = \frac{a_i^2 + b_i^2}{c_i}.$$
We have the lower estimate for $S(a_{i+1}, b_{i+1}, c_{i+1})$:
$$\begin{align*} S(a_{i+1}, b_{i+1}, c_{i+1}) &= \frac{b_i^2 + c_i^2}{a_i} + \frac{c_i^2 + a_i^2}{b_i} + \frac{a_i^2 + b_i^2}{c_i} \ge \frac{2b_i c_i}{a_i} + \frac{2c_i a_i}{b_i} + \frac{2a_i b_i}{c_i} = \\ &= \frac{b_i c_i}{a_i} + \frac{b_i c_i}{a_i} + \frac{c_i a_i}{b_i} + \frac{c_i a_i}{b_i} + \frac{a_i b_i}{c_i} + \frac{a_i b_i}{c_i} = \\ &= \left( \frac{b_i c_i}{a_i} + \frac{a_i b_i}{c_i} \right) + \left( \frac{b_i c_i}{a_i} + \frac{c_i a_i}{b_i} \right) + \left( \frac{a_i b_i}{c_i} + \frac{c_i a_i}{b_i} \right) = \\ &= b_i \left( \frac{c_i}{a_i} + \frac{a_i}{c_i} \right) + c_i \left( \frac{b_i}{a_i} + \frac{a_i}{b_i} \right) + a_i \left( \frac{b_i}{c_i} + \frac{c_i}{b_i} \right) \ge \left[ \frac{\alpha}{\beta} + \frac{\beta}{\alpha} \ge 2 \right] \ge \\ &\ge 2(a_i + b_i + c_i) = 2S(a_i, b_i, c_i). \end{align*}$$
Therefore, $S(a_i, b_i, c_i) \le \frac{1}{2} S(a_{i+1}, b_{i+1}, c_{i+1})$ for all $i = 0, 1, 2, 3, 4$. Hence,
$$\begin{align*} S(a_0, b_0, c_0) &\le \frac{1}{2} S(a_1, b_1, c_1) \le \frac{1}{2^2} S(a_2, b_2, c_2) \le \dots \le \frac{1}{2^5} S(a_5, b_5, c_5) = \\ &= \frac{2016}{32} = 63. \end{align*}$$
Thus, the sum of the initial numbers on the blackboard is less than or equal to 63.
The following example shows that the sum of the initial numbers on the blackboard can be equal to 63. Let $a_0 = b_0 = c_0 = 21$, then
$$\begin{align*} (a_0, b_0, c_0) = (21, 21, 21) &\longrightarrow (42, 42, 42) = (2 \cdot 21, 2 \cdot 21, 2 \cdot 21) &\longrightarrow \\ &\longrightarrow (2^2 \cdot 21, 2^2 \cdot 21, 2^2 \cdot 21) &\longrightarrow \dots &\longrightarrow (2^5 \cdot 21, 2^5 \cdot 21, 2^5 \cdot 21) = (a_5, b_5, c_5), \end{align*}$$
and the sum
$$S(a_5, b_5, c_5) = S(2^5 \cdot 21, 2^5 \cdot 21, 2^5 \cdot 21) = 2^5 S(21, 21, 21) = 32 \cdot 63 = 2016.$$