Thai Mathematical Olympiad

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 $$a_n=\{\begin{array}{ll}1& \text{if }n\equiv 1,2,3(\mathrm{mod}6),\\ 2& \text{if }n\equiv 4,5,6(\mathrm{mod}6).\end{array}$$ 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.