Solution:
We prove by induction that $o(n) = 2^{2n-1} - 2^{n-1}$. For $n = 1$, this reads $o(1) = 2^{1} - 2^{0} = 1$, which is obviously true—the only such $2$-tuple is $(1, 1)$. Suppose it is true for $n$.

If $(a_{1}, a_{2}, \ldots, a_{n}, b_{1}, b_{2}, \ldots, b_{n})$ gives an odd sum, then we can take $(a_{n+1}, b_{n+1})$ to be any of $(0, 0), (0, 1), (1, 0)$ and still get an odd sum for $(a_{1}, a_{2}, \ldots, a_{n+1}, b_{1}, b_{2}, \ldots, b_{n+1})$. On the other hand, if $(a_{1}, a_{2}, \ldots, a_{n}, b_{1}, b_{2}, \ldots, b_{n})$ is even, then we must have $a_{n+1} = b_{n+1} = 1$ to get an odd sum. Thus $o(n+1) = 3 o(n) + e(n)$.

But $o(n) = 2^{2n-1} - 2^{n-1}$ and $e(n) = (o(n) + e(n)) - o(n) = 2^{2n} - (2^{2n-1} - 2^{n-1}) = 2^{2n-1} + 2^{n-1}$.

So $o(n+1) = 3 o(n) + e(n) = 3(2^{2n-1} - 2^{n-1}) + (2^{2n-1} + 2^{n-1}) = 4 \cdot 2^{2n-1} - 2 \cdot 2^{n-1} = 2^{2(n+1)-1} - 2^{(n+1)-1}$, which establishes the result for $n+1$ and hence for all $n$.

Hence $e(n) = 2^{2n} - o(n) = 2^{2n-1} + 2^{n-1}$ and $\dfrac{o(n)}{e(n)} = \dfrac{2^{n} - 1}{2^{n} + 1}$.