From $a_n a_{n+3} = a_{n+2} a_{n+5}$ for every positive integer $n$, we have
$a_{n+1} a_{n+4} = a_{n+3} a_{n+6}$ and $a_{n+2} a_{n+5} = a_{n+4} a_{n+7}$. Then
$$a_n a_{n+3} \cdot a_{n+1} a_{n+4} \cdot a_{n+2} a_{n+5} = a_{n+2} a_{n+5} \cdot a_{n+3} a_{n+6} \cdot a_{n+4} a_{n+7}.$$
Therefore, $a_n a_{n+1} = a_{n+6} a_{n+7}$ for every positive integer $n$. Thus,
$$\sum_{k=1}^{2550} a_{2k} a_{2k-1} = \frac{2550}{3} (a_1 a_2 + a_3 a_4 + a_5 a_6) = 850 (a_1 a_2 + a_3 a_4 + a_5 a_6)$$
We will show that 850 is the largest positive integer that always divides $\sum_{k=1}^{2550} a_{2k} a_{2k-1}$. Consider the sequence $\{a_n\}_{n \ge 1}$ defined by
$$a_n = \begin{cases} 1 & \text{if } n \equiv 1, 2, 3 \pmod 6, \\ 2 & \text{if } n \equiv 4, 5, 6 \pmod 6. \end{cases}$$
It can be seen that $a_n a_{n+3} = 2 = a_{n+2} a_{n+5}$ for every positive integer $n$; so, it satisfied the condition in the problem and
$$\sum_{k=1}^{2550} a_{2k} a_{2k-1} = 850(1 \cdot 1 + 1 \cdot 2 + 2 \cdot 2) = 850 \cdot 7.$$
Consider another sequence $\{a_n\}_{n \ge 1}$ defined by $a_n = 1$ for every positive integer $n$. We can see that $a_n a_{n+3} = 1 = a_{n+2} a_{n+5}$ for every positive integer $n$, and
$$\sum_{k=1}^{2550} a_{2k} a_{2k-1} = 850(1 \cdot 1 + 1 \cdot 1 + 1 \cdot 1) = 850 \cdot 3.$$
Thus, 850 is the desired largest integer.