Given a sequence $$a_1, a_2, a_3, \ldots$$, where $$a_n = a_{n-1} - a_{n-2}$$, for $$n \in \mathbb{N}$$ and $$n \ge 3$$. Also, $$S_n = a_1 + a_2 + a_3 + \cdots + a_{n-1} + a_n$$. If $$S_{2003} = 2026$$ and $$S_{2026} = 2003$$, then find $$S_{2030}$$.

Let $$a_1 = x$$ and $$a_2 = y$$. Then $$a_3 = y - x$$, $$a_4 = -x$$, $$a_5 = -y$$, $$a_6 = x - y$$, $$a_7 = x$$ and $$a_8 = y$$, so the sequence has period 6, any 6 consecutive terms sum to 0, and $$S_{n+6} = S_n$$. Since $$2003 = 6 \times 333 + 5$$, $$S_{2003} = S_5 = y - x = 2026$$, and since $$2026 = 6 \times 337 + 4$$, $$S_{2026} = S_4 = 2y - x = 2003$$. Subtracting gives $$y = 2003 - 2026 = -23$$ and then $$x = y - 2026 = -2049$$. Finally $$2030 = 6 \times 338 + 2$$, so $$S_{2030} = S_2 = x + y = -2049 - 23 = -2072$$.