The prime factorisation of $2020$ is $2020 = 4 \cdot 5 \cdot 101$ and so a complete list of divisors of $2020$ is:
$$\begin{aligned} d_1 &= 1,\quad d_2 = 2,\quad d_3 = 4,\quad d_4 = 5,\quad d_5 = 10,\quad d_6 = 20,\quad d_7 = 101, \\ d_8 &= 202,\quad d_9 = 404,\quad d_{10} = 505,\quad d_{11} = 1010,\quad d_{12} = 2020. \end{aligned}$$
The required properties, $2020 = d_3d_4d_7$ and $d_3d_4 < d_7$, are now easy to check.

As $N = d_3d_4d_7$, the product $d_3d_4$ is a divisor of $N$, which was assumed to be smaller than $d_7$. Hence $d_3d_4 = d_5$ or $d_3d_4 = d_6$. We first exclude that $d_3d_4 = d_5$. In this case, $N = d_3d_4d_7 = d_5d_7$ and the number of divisors of $N$ is equal to $11$. Because the number of positive divisors of $N = \prod p_i^{e_i}$ is equal to $\prod (e_i + 1)$, any number with exactly $11$ divisors must be of the form $N = p^{10}$ where $p$ is a prime number. But then $d_3 = p^2$, $d_4 = p^3$ and $d_5 = p^4$ and so $d_3d_4 \neq d_5$. This shows that we must have $d_3d_4 = d_6$. Then $N = d_3d_4d_7 = d_6d_7$ and $N$ has $12$ divisors.
As $12 = 6 \cdot 2 = 4 \cdot 3 = 3 \cdot 2 \cdot 2$ we have to consider the following four cases: $N = p^{11}$, $N = p^5q$, $N = p^3q^2$, $N = p^2qr$ where $p, q, r$ are distinct primes.

$N = p^{11}$: In this case, $d_3 = p^2$, $d_4 = p^3$ and $d_7 = p^6$, and we see that $N = d_3d_4d_7$ as well as $d_3d_4 < d_7$ as required.

$N = p^5q$: We have $1 < p < p^2 < p^3 < p^4 < p^5$ and we can order the divisors of $N$ once we know the size of $q$. There are six possibilities

and we easily check that $d_3d_4 < d_7$ holds when $1 < q < p$ and when $p^5 < q$.

$N = p^3q^2$: We will investigate all possibilities for $d_6d_7 = p^3q^2$ and check if $d_3d_4 = d_6$. The pairs $\{d_i, d_{13-i}\}$, whose product is $N$, are the following:
$$\{1, p^3q^2\},\quad \{p, p^2q^2\},\quad \{p^2, pq^2\},\quad \{p^3, q^2\},\quad \{q, p^3q\},\quad \{pq, p^2q\}.$$
Because $p^3q^2$, $p^2q^2$ and $p^3q$ have more than $7$ factors, these cannot be equal to $d_6$ or $d_7$, so we have only three options to consider for $\{d_6, d_7\}$, namely $\{p^2, pq^2\}$, $\{p^3, q^2\}$ or $\{pq, p^2q\}$.
If $\{d_6, d_7\} = \{p^2, pq^2\}$, then $d_6 = pq^2$ and $d_7 = p^2$, because we cannot have $d_6 = p^2 = d_3d_4$ as $d_3 > 1$. From $pq^2 = d_6 < d_7 = p^2$, we get $q^2 < p$. Hence, the divisors of $N$ need to satisfy
$$1 < q < q^2 < p < pq < pq^2 < p^2 < p^2q < p^2q^2 < p^3 < p^3q < p^3q^2.$$
We see now that $d_3d_4 = d_6$ and we get a working solution.

If $\{d_6, d_7\} = \{p^3, q^2\}$, then $d_6 = p^3$ and $d_7 = q^2$, because we cannot have $d_6 = q^2 = d_3d_4$ as $d_3 > 1$. The equation $p^3 = d_6 = d_3d_4$ can only be achieved with $d_3 = p$ and $d_4 = p^2$. This implies $d_2 = q$ and so $1 < q < p < p^2$. But then $d_3 = p < pq < p^2 = d_4$, a contradiction.
If $\{d_6, d_7\} = \{pq, p^2q\}$, then $d_6 = pq < p^2q = d_7$, and $pq = d_6 = d_3d_4$ could only be achieved with $d_3 = p, d_4 = q$ or $d_3 = q, d_4 = p$. Both are impossible as there would be no option left for $d_2$.

$N = p^2qr$: We may assume that $1 < q < r < qr$. Because $p^2q$ has six divisors and all other divisors of $N$ are multiples of $r$, $d_k = r$ for some $3 \le k \le 7$.
If $d_7 = r$, then $d_6 = p^2q$ and we automatically have $d_6 = d_3d_4$ as $p^2q$ has six divisors. This case occurs precisely when $p^2q < r$.
If $d_6 = r$, then $d_6 = d_3d_4$ is impossible as $d_3 > 1$.
If $d_5 = r$, then $d_8 = p^2q$ and $d_6, d_7$ must form one of the pairs $\{p, pqr\}$, $\{pq, pr\}$, $\{p^2, qr\}$. The first of these three is impossible, as $p < pq < pr < pqr$. Secondly, when $d_6 = pq < pr = d_7$, $pq = d_6 = d_3d_4$ could only be achieved with $d_3 = p, d_4 = q$ or $d_3 = q, d_4 = p$. Both are impossible as there would be no option left for $d_2$. For the third option we first note that $d_6 = p^2$ can be ruled out as before using $d_6 = d_3d_4$. However, $d_6 = qr$ is impossible as well, because $d_5 = r$ and we again get a problem with $d_6 = d_3d_4$.
If $d_4 = r$, then $d_2 = p$ or $d_3 = p$. In the first case we would have $1 < p < q < r$ and so $d_6 = d_3d_4 = qr$, which implies $d_7 = p^2$. However, we have $p^2 < pq < pr$ which would mean that both members of the pair $\{pq, pr\}$ are larger than $d_7$, contradiction. In the second case, we would have $1 < q < p < r$ and so $d_6 = d_3d_4 = pr$, which implies $d_7 = pq$. But $q < r$ implies $d_7 = pq < pr = d_6$, contradiction.
If $d_3 = r$, then $d_4 = p$ because $d_6 = d_3d_4$ cannot be divisible by $r^2$. We get $d_6 = pr$ and $d_7 = pq$. But $q < r$ implies $d_7 = pq < pr = d_6$, a contradiction.

To summarise, the complete list of solutions is: