Define $d(i, j) = (a_i - a_j)^2 + (b_i - b_j)^2$. Notice that
$$(a_n - a_{n-1})^2 + (a_n - a_{n-2})^2 - (a_{n-1} - a_{n-2})^2 = 2(a_n - a_{n-1})(a_n - a_{n-2})$$
and
$$(b_n - b_{n-1})^2 + (b_n - b_{n-2})^2 - (b_{n-1} - b_{n-2})^2 = 2(b_n - b_{n-1})(b_n - b_{n-2}).$$
Adding these two equations and using the given implies that
$$d(n, n-1) + d(n, n-2) - d(n-1, n-2) = 0,$$
hence
$$d(n, n-1) = d(n-1, n-2) - d(n, n-2) \le d(n-1, n-2),$$
where equality holds if and only if $a_n = a_{n-2}$ and $b_n = b_{n-2}$. Therefore, the sequence
$$\{d(n, n-1)\}_{n=2}^{\infty} = (d(2, 1), d(3, 2), d(4, 3), \dots)$$
is nonincreasing. Since all the terms of this sequence are nonnegative integers, the sequence must eventually become constant: $d(n, n-1) = d(n-1, n-2)$ for all sufficiently large $n$, which in particular implies $a_n = a_{n-2}$ for all sufficiently large $n$. So whenever $n$ is large enough, we have
$$a_{n+2008} = a_{n+2006} = a_{n+2004} = \dots = a_n.$$

(Based on work by Palmer Mebane) Let $x_n = a_n - a_{n-1}$ and $y_n = b_n - b_{n-1}$. Then the given equation reads
$$x_n(x_n + x_{n-1}) + y_n(y_n + y_{n-1}) = 0.$$
To see this, note that $0 \le (x_{n-1} + x_n)^2 + (y_{n-1} + y_n)^2$, with equality if and only if $x_n = -x_{n-1}$ and $y_n = -y_{n-1}$. Expanding and using the given equation we obtain
$$0 = 2x_n(x_n + x_{n-1}) + 2y_n(y_n + y_{n-1}) \le (x_{n-1} + x_n)^2 + (y_{n-1} + y_n)^2$$
or
$$x_n^2 + y_n^2 \le x_{n-1}^2 + y_{n-1}^2,$$
with equality if and only if $x_n = -x_{n-1}$ and $y_n = -y_{n-1}$. Thus we have the claim.

Therefore, because $x_n^2 + y_n^2 \ge 0$ is an integer, by the well-ordering principle, $x_n^2 + y_n^2 < x_{n-1}^2 + y_{n-1}^2$ can only hold finitely many times. Thus there is a $k$ such that for all $n > k$, $x_n = -x_{n-1}$ and $y_n = -y_{n-1}$. This gives us the desired $k$ such that $a_k = a_{k+2008}$.