2003 is prime, so $a^{2002} = 1 \bmod 2003$ for any $a$ not divisible by 2003. Thus $a_{n+1} = a_{n}^{-1} + b_{n} \bmod 2003, b_{n+1} = b_{n}^{-1} + a_{n} \bmod 2003$. Put $c_{n} = a_{n} b_{n}$. Then $c_{n+1} = c_{n} + \frac{1}{c_{n}} + 2 = \frac{(c_{n} + 1)^2}{c_{n}} \bmod 2003$. So if $c_{n} \neq 0 \bmod 2003$, then $c_{n+1} \neq 0 \bmod 2003$ unless $c_{n} = -1 \bmod 2003$. Then if $c_{n+1} = -1 \bmod 2003$, we must have $\frac{c_{n}^2 + 3c_{n} + 1}{c_{n}} = 0 \bmod 2003$, so $c_{n}^2 + 3c_{n} + 1 = 0 \bmod 2003$. Note that $c_{0} = 4$. So it is sufficient to show that there are no solutions to $x^2 + 3x + 1 = 0 \bmod 2003$, or equivalently to $(x - 1000)^2 = 1000^2 - 1 = 502 \bmod 2003$. In other words, we have to show that 502 is a quadratic non-residue mod 2003.

The easiest way to do that is to use the law of quadratic reciprocity, but that is almost certainly outside the syllabus. We note that $502 = 5 \bmod 2003$, so 502 is a square iff 5 is a square. It is sufficient to show that $5^{1001} = -1 \bmod 2003$, for then if we had $x^2 = 5$, we would have $x^{2002} = -1 \bmod 2003$, whereas we know that $x^{2002} = 1 \bmod 2003$. We note that $1001 = 7 \cdot 11 \cdot 13$. We start by showing that $5^7 = 8 \bmod 2003$. We have $5^5 = 3125 = 1122 \bmod 2003$, so $5^6 = 5610 = 1604 \bmod 2003$, so $5^7 = 8020 = 8 \bmod 2003$.

We calculate successively $2^{11} = 2048 = 45 \bmod 2003$, so $2^{22} = 2025 = 22 \bmod 2003$. Multiplying by 22 is relatively easy, so $2^{44} = 484$, $2^{66} = 10648 = 633$, $2^{88} = 13926 = -95$, $2^{110} = -2090 = -87$, $2^{132} = -1914 = 89$, $2^{143} = 4005 = -1$ all $\bmod 2003$. Hence $8^{11 \cdot 13} = -1 \bmod 2003$, so $5^{1001} = -1 \bmod 2003$, as required, and we are done.