Solution:
We claim that the maximum value of $ac$ is $2 \times 505^{2} = 510050$, and this is uniquely achieved by $(a, b, c, d) = (1010, 506, 505, -2021)$.

To prove this we start by rearranging the expression $\frac{a + b}{b + c} = \frac{c + d}{d + a}$ to get $(c + b)(c + d) = (a + b)(a + d)$. Now expand the brackets and move all the terms to one side to get $(c - a)(a + b + c + d) = 0$. Since $c > a$ we can divide by $c - a$ to get
$$a + b + c + d = 0.$$
Note that this is equivalent to the $\frac{a + b}{b + c} = \frac{c + d}{d + a}$ condition. Now assume that $(a, b, c, d)$ is the tuple such that $ac$ is maximised. Since $4a > a + b + c + d = 0$ we trivially have $a > 0$. Furthermore if $c \leq 0$ then we would have $ac \leq 0$ which would contradict the definition that $ac$ is maximal. Therefore
$$a > b > c > 0 \geq d \geq -2021.$$
Now we perform two little proofs by contradiction:
- If $d > -2021$ then we could consider $(a', b', c', d') = (a + 1, b, c, d - 1)$ so that
$$a'c' = (a + 1)c > ac.$$
However this would contradict the definition that $ac$ was maximal.
Therefore $d = -2021$
- If $b > c + 1$ then we could consider $(a', b', c', d') = (a + 1, b - 1, c, d)$ so that
$$a'c' = (a + 1)c > ac.$$
However this would contradict the definition that $ac$ was maximal.
Therefore $b = c + 1$
Therefore $(a, b, c, d) = (a, c + 1, c, -2021)$, and since $a + b + c + d = 0$, this means $a = 2020 - 2c$. So the final expression we are trying to maximise is
$$a c = (2020 - 2c)c$$
$$\qquad = 2\times 505^{2} - 2(c - 505)^{2}$$
$$\qquad \geq 2\times 505^{2}.$$
with equality iff $c = 505$. Therefore the maximum value of $ac$ is $2 \times 505^{2} = 510050$ which is achieved by
$$(a, b, c, d) = (1010, 506, 505, -2021).$$