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$c_n$ must be zero for all odd $n$. Proof: WLOG suppose that $a_1 \geq a_2 \geq ... \geq a_8$. If $a_1+a_8 > 0$ then for sufficiently high odd $n$, $c_n$ will be dominated by $a_1$ alone i.e. it will always be positive. Similarly if $a_1+a_8 < 0$; hence $a_1=-a_8$. Now for odd $n$ these terms cancel, so we can repeat for the remaining values. Now all the terms cancel for all odd $n$. Since some $a_i$ is nonzero $c_n > 0$ for even $n$.

QED Many early IMO problems are simply jokes in comparing with present problems.

$n$ can't be even since that would mean that all $a_1,...,a_8=0$ which goes against the conditions. If n is even. Now we must prove that for all odd n the statement is true. Assume the opposite that: $a_1>=a_2>=.....>=a_8$ Case 1: $a_1+a_8>0$ Then we would have that $c_n$ is infinite for huge n and then there wouldn't be infinitely many $c_n$ which are equal to 0. Case 2: $a_1+a_8<0$ Then we would have that $c_n$ is infinite for huge n and then there wouldn't be infinitely many $c_n$ which are equal to 0. Therefore, $a_1+a_8=0$. Now we can repeat this process for the pairs.