Suppose that $2003|a_k$ or $b_k$ for some $k$ and that $k$ is the smallest index among these two sequences such that $2003$ divides one term. Wlog we can assume that $2003|a_k$. As 2003 is prime we can rewrite our recurrences for $a_n,b_n$ not divisible by 2003 as $a_{n+1} \equiv \frac{1}{a_n}+b_n$ and $b_{n+1} \equiv \frac{1}{b_n} +a_n$ which leads to $(a_{n+1}-b_n)a_n \equiv 1$ and $(b_{n+1}-a_n)b_n \equiv 1$ (everything modulo 2003 of course). Now $a_k \equiv 0$ implies $b_{k-1}a_{k-1} \equiv -1$ (*). We get also that for $l < k$, $a_l a_{l-1} \equiv b_l b_{l-1}$. So $\frac{a_l}{b_l} \equiv 4$ or $\frac{1}{4}$. These leads with (*) to $a_{k-1} ^2 \equiv -4$ or $-\frac{1}{4}$ - contradiction in both cases. Hence 2003 does not divide any term of these two sequences

Use Induction. We can easily check that $a_{0}$ and $a_{1}$ aren't divisible by 2003. Let us assume $2003$ doesn't divide $a_{k-1}$ and $a_{k-2}$. For sake of contradiction, assume that $2003|a_{k}$. Then \begin{eqnarray}a_{k}&\equiv& a_{k-1}^{2001}+b_{k-1}\equiv 0\left(\mod 2003\right) \nonumber\\ a_{k-1}a_{k}&\equiv&a_{k-1}^{2002}+a_{k-1}b_{k-1}\equiv 0\left(\mod 2003\right)\nonumber\\ &\equiv& 1+a_{k-2}^{2002}+b_{k-2}^{2002}+\left(a_{k-2}b_{k-2}\right)^{2001}+\left(a_{k-1}b_{k-1}\right)\equiv 0\left(\mod 2003\right)\nonumber\\ xa_{k-1}a_{k}&\equiv& x^{2}+3x+1\equiv 0\left(\mod 2003\right)\nonumber \end{eqnarray} where $x=a_{k-2}b_{k-2}$. Thus, there exists an integer $k$ such that \[x^{2}+3x+1=2003k\] but $x$ is an integer so $9-4(1-2003k)$ must be a perfect square, but that implies 5 is a square modulo 2003. By Quadratic Reciprocity, \[\left(\frac{5}{2003}\right)\left(\frac{2003}{5}\right)=\left(-1\right)^{\frac{5-1}{2}\cdot\frac{2003-1}{2}}=1\] and $\left(\frac{2003}{5}\right)=\left(\frac{3}{5}\right)=-1$, so $\left(\frac{5}{2003}\right)=-1$, contradiction. Thus, 2003 doesn't divide $a_{k}$. Same arguments can be used to prove 2003 doesn't divide $b_{k}$.

Assume for contradiction that $2003$ is the divisor of some term in one of the sequences. Let $k$ be the minimal positive integer such that $2003$ divides one of $\{ a_k, b_k \}$ where $k \ge 1$. We will now prove by induction that there exists a positive integer $x_i$ such that $x_{i}^{2} \equiv a_i \cdot b_i \pmod{2003}$ for all $0 \le i \le k-1$. Note that the claim holds for $i=0$ and now assume that the claim holds for a certain $i$ where $0 \le i \le k-1$. Now note that since $2003$ is prime, \[x_{i}^2 \cdot a_{i+1} \cdot b_{i+1} \equiv (a_{i}^{2002} + a_i \cdot b_i) \cdot (b_{i}^{2002} + a_i \cdot b_i) \equiv (a_i \cdot b_i + 1)^2 \pmod{2003}.\] Since $k$ is the minimal positive integer such that $2003$ divides one of $\{ a_k, b_k \}$, it follows that $2003$ does not divide either $a_i$ or $b_i$ since $0 \le i \le k-1$. Therefore $2003$ does not divide $x_i$ and hence there exists a positive integer $m$ such that $m\cdot x_i \equiv 1 \pmod{2003}$ since $2003$ is prime. Hence \[a_{i+1} \cdot b_{i+1} \equiv m^2 \cdot x_{i}^2 \cdot a_{i+1} \cdot b_{i+1} \equiv \left( m(a_i \cdot b_i + 1) \right)^2 \pmod{2003}.\] This completes the induction. Now if $2003$ divides $a_k$, it follows that \[0 \equiv a_{k-1}a_k \equiv a_{k-1}^{2002} + a_{k-1} b_{k-1} \equiv 1 + x_{k-1}^2 \pmod{2003}.\] However, since $2003 \equiv 3 \pmod{4}$, this is a contradiction since $-1$ is not a quadratic residue modulo $2003$. A similar contradiction arises if $2003$ divides $b_k$. Hence no term in either sequence is divisible by $2003$.

(All congruences are modulo $2003$.) Suppose to the contrary, let $n_0$ be the minimal such $n$, then $\forall n \in \{ 1, \cdots , n_0-1 \}$ $gcd ( a_n , 2003) = gcd (b_n, 2003) = 1$ (as $2003$ is prime) whence by Fermat's Little Theorem, $\forall m \leq n_0,$ $$a_{m} \equiv a_{m-1}^{-1} + b_{m-1} \pmod{2003}, b_{m} \equiv a_{m-1} + b_{m-1}^{-1} \pmod{2003} \hspace{1 mm} \cdots (1)$$$$\implies 1 \equiv b_{m-1} b_{m-1}^{-1} \equiv (a_m-a_{m-1}^{-1})(b_m-a_{m-1}) \iff a_mb_m \equiv a_ma_{m-1}+b_ma_{m-1}^{-1} \hspace{1mm} (\forall m \leq n_0) \cdots (2)$$ Now we claim that $a_{n_0} \equiv b_{n_0} \equiv 0 \pmod{2003}$. To show this we insert $m:=n_0$ into $(2)$ to obtain $a_{n_0}a_{n_0-1} \equiv -b_{n_0}a_{n_0-1}^{-1}$. Now since $2003$ is prime $2003 | a_{n_0}b_{n_0}$ $\implies 2003|a_{n_0}$ or $2003| b_{n_0}$. This means both sides of the previous congruence are $0$ and by the minimality of $n_0$ (if $2003 | a_{n_0}$ then $b_{n_0}a_{n_0-1}^{-1} \equiv 0 \implies b_{n_0} \equiv 0$ as $a_{n_0-1} \in U_{2003}$ and if $2003| b_{n_0}$, then $2003| a_{n_0}a_{n_0-1} \implies 2003 | a_{n_0}$ otherwise $2003| a_{n_0-1}b_{n_0-1}$ contradicts the minimality of $n_0$), the claim follows. Next we observe that inserting $m:=n_0$ into $(1)$ we get $a_{n_0-1} b_{n_0-1} \equiv -1 ...(3)$ whence employing $(2)$ with $m := n_0-1$ we obtain $$-1 \equiv a_{n_0-1}b_{n_0-1} \equiv a_{n_0-1}a_{n_0-2}+b_{n_0-1} a_{n_0 - 2}^{-1} \equiv a_{n_0-1}a_{n_0-2} - (a_{n_0-1}a_{n_0-2})^{-1} \pmod{2003}$$$$ \iff (a_{n_0-1}a_{n_0-2})^2 + a_{n_0-1}a_{n_0-2} -1 \equiv 0 \iff (2a_{n_0-1}a_{n_0-2} + 1)^2 \equiv 5 \pmod{2003}$$ This means $5 \equiv 1 \pmod{4}$ is a quadratic residue modulo $2003$, so that by the Law of Quadratic Reciprocity, $3 \equiv 2003 \pmod{5}$ is also a quadratic residue modulo $5$, an immediate contradiction! This completes the proof.

We rearrange the definitions to become $a_{n+1} = b_n^{2001}+a_n = \frac{a_nb_n+1}{b_n}$ and $b_{n+1} = a_n^{2001}+b_n = \frac{a_nb_n+1}{a_n}$. We claim that for all $n$ we have $b_n \equiv 4a_n$. The base case is given. Then, otherwise we have \[\frac{b_{n+1}}{a_{n+1}} = \frac{a_n^{2001}+b_n}{b_n^{2001}+a_n}\equiv \frac{b_n+a_nb_n^2}{a_n+a_n^2b_n}=\frac{b_n}{a_n}=4 \pmod{2003}\]AFTSOC that at some point we have $a_nb_n+1$ is divisible by 2003. Consider the first time this happens. Then, \[-1\equiv a_nb_n \equiv 4a_n^2 \pmod{2003}\]This implies that -1 is a quadratic residue mod 2003, but since 2003 is 3 mod 4, this is a contradiction, so we will never have $2003\mid a_nb_n+1$, so $a_{n+1},b_{n+1}$ will never be divisible by 2003.

We work modulo the prime $2003$. If $a_n \not \equiv 0$ then we have $a_{n+1}=a_n^{-1}+b_n$ and likewise if $b_n \not \equiv 0$ we have $b_{n+1}=b_n^{-1}+a_n$, hence $a_{n+1}b_{n+1}=a_nb_n+(a_nb_n)^{-1}+2$. Thus if some minimal $k$ has $a_kb_k=0$ we must have $a_{k-1}b_{k-1}$ be a root of $x^2+2x+1$ in $\mathbb{F}_{2003}$, hence $a_{k-1}b_{k-1} \equiv 1$. But then $a_{k-2}b_{k-2}$ is a root of $x^2+3x+1$ in $\mathbb{F}_{2003}$, implying that $3^2-4(1)=5$ is a quadratic residue modulo $2003$ by the quadratic formula. On the other hand, by quadratic reciprocity we have $$1=\left(\frac{5}{2003}\right)\left(\frac{2003}{5}\right)\left(\frac{5}{2003}\right)\left(\frac{3}{5}\right)=-\left(\frac{5}{2003}\right),$$so $5$ is not a quadratic residue. $\blacksquare$

The problem more or less implies the solution here. Observe that $a_n^{2001} \equiv \frac 1{a_n} \pmod {2003}$, so we can rewrite the recurrence relations as \begin{align*} a_na_{n+1} &= 1+b_na_n \\ b_nb_{n+1} &= 1 + b_na_n. \end{align*}In particular, $a_na_{n+1} = b_nb_{n+1}$ and $a_{n+1}a_{n+2} = b_{n+1}b_{n+2}$, so $\frac{a_{n+2}}{a_n} = \frac{b_{n+2}}{b_n}$ for each $n$. This means that $a_{n+2}b_{n+2} = k^2 (a_nb_n)$ for each $n$ for some $k \in \mathbb F_{2003}$, i.e. $\left(\frac{a_nb_n}{2003}\right)$ is invariant for each parity class. It suffices to show that we cannot have $b_na_n = -1$ at some point. But this is clear as $a_0b_0 = 4$ and $a_1b_1 = \frac{25}4$ are both quadratic residues mod $2003$, whilst $\left(\frac{-1}{2003}\right) = -1$.