Divisibility - parity problem

$s(n)=\frac{1}{6}n^3-\frac{1}{2}n^2+\frac{1}{3}n=\frac{1}{6}(n-2)(n-1)n=\frac{1}{6}t(n)$

$s(n)$ is divisible by 4 if and only if $t(n)$ is divisible by 8. This is the case for all even numbers $n$ , because then one of the numbers $n-2$ and $n$ is divisible by 4. (Every other number even number is divisible by 4, and the other divisible by 2). There exists 1004 even numbers in the relevant area. If n is an odd number, $n-1$ must be divisible by 8. This is the case for $\frac{2008}{8}=251$ of the odd numbers in the relevant area. Thus, in total there exists $1004+251=1255$ numbers $n$ such that $0<n\leq2008$ and $s(n)$ is divisible by 4.