Let $b=k-a$. Then the equation is equivalent with $$(3k+2020)a^2-(3k^2+2020k)a+k^3-8080=0$$It is quadric to $a$ and it's discriminant is $$D=(3k+2020)(-k^3+2020k^2+32320)$$Obviously $D=>0$. Therefore $2020k^2+32320=>k^3$. But this is true for finitely many $k$.... Just check them

Wrong solution

We will show the following theorem. Theorem Let $k$ be a product of distinct primes congruent to $5\bmod 6$. For $( a,b) \in \mathbb{N}^{2}$, $\frac{a^{3} +b^{3}}{ab+4} =4k$ if and only if $\{a,b\} =\{2k+1,2k-1\}$.

Are you guys kidding? This is a P1, so chances are it must admit a simple solution. Here is a simpler one. I claim $(1011,1009)$ and $(1009,1011)$ are the only solutions. Note that $2020=4\cdot 5 \cdot 101$, and keep in mind that $101$ and $5$ are both primes of form $6\ell-1$. In particular, for such primes $p$, it is very well-known that $-3$ is not a quadratic residue modulo $p$. Equipped with this fact, we now attack the problem. We have $$ (a+b)(a^2-ab+b^2)=2020(ab+4). $$Observe that if $101\mid a$, then $101\mid b$ as well, but since $v_{101}(2020)=1$, this is not possible. Thus, $101\nmid a,b$. Similarly, $5\nmid a,b$. Suppose now that $101\mid a^2-ab+b^2 \Rightarrow 101\mid (2a-b)^2+3b^2$. Since $101\nmid b$, it then follows that $-3$ is a quadratic residue modulo $101$, not possible. Thus, $101\mid a+b$; and similarly, $5\mid a+b$. This means $a^2-ab+b^2 $ is of form $x_1 y_1$, where $y_1$ is a divisor of $ab+4$, and $x_1\mid 4$. Furthermore, $a$ and $b$ clearly have the same parity. This yields $$ a^2-ab+b^2\mid 2(ab+4). $$Now, two cases to consider: if $a,b$ are both odd, so do $a^2-ab+b^2$. Hence, $a^2-ab+b^2\mid ab+4$. In particular, $(a-b)^2\leqslant 4$. Now clearly $a\neq b$, and furthermore, $a-b=1$ is impossible since both are odd. Thus, if $a$ and $b$ are both odd, then $|a-b|=2$. Inserting this back to the equation, we immediately conclude that $(a,b)=(1011,1009)$ and $(1009,1011)$ are solutions. Now suppose $a,b$ are both even, and let $a=2a_1$ and $b=2b_1$. Then, $a_1^2-a_1b_1+b_1^2\mid 2a_1b_1+2$. Now, if $a_1,b_1$ are both even, then $4\mid a_1^2-a_1b_1+b_1^2$, but $2\mid\mid 2a_1b_1+2$. Thus, either $a_1$ and $b_1$ are both odd, or one of them is even; and the other is odd. In both cases; $a_1^2-a_1b_1+b_1^2$ is odd, hence $a_1^2-a_1b_1+b_1^2\mid a_1b_1+1$, that is, $|a_1-b_1|=1$. Inserting this back, however, we immediately conclude that this brings no solutions. Hence, $(1011,1009)$ and $(1009,1011)$ are the only solutions, as claimed earlier.

We have $$\frac{(a+b)(a^2-ab+b^2)}{ab+4}=2020=2^2.5.101\Rightarrow (a+b)(a^2-ab+b^2)=2^2.5.101.(ab+4)$$ Now, If $5\mid a^2-ab+b^2$, since $5\neq 12k\pm1$, we know that $-3$ is not a quadratic residue in modulo $5$. Then it follows that $a^2-ab+b^2\equiv0\rightarrow \frac({2a-b}{b})^2\equiv-3\mod5$, if $5\nmid a,b$ and obviously, its impossible. $\Rightarrow 5\mid a,b$ The result here is that we must have $5\mid a+b$. Same applies for $101$ too. Finally we get $505\mid a+b$. Wlog let $a\geq b$. If $a-b=0$, or $a=b$, we have $$\frac{2a^3}{a^2+4}\in\mathbb Z\Rightarrow 2a.(a^2)\equiv 2a.(-4)\equiv-8a\mod a^2+4\Rightarrow a^2+4\mid 8a\Rightarrow a\leq 8$$But this gives $a+b=2a\leq16$, which is not divisible by $505$. If $a-b=1$, you can see it is not possible by doing the same thing. If $a-b\geq2$, we have $$2020=(a+b)\frac{a^2-ab+b^2}{ab+4}=(a+b)\frac{ab+(a-b)^2}{ab+4}\geq (a+b)\frac{ab+4}{ab+4}=a+b$$ So we get $505\mid a+b$, $a+b\leq2020$, So there are $4$ cases we need to investigate, let $a+b=505k.\Rightarrow$ $$4=k.\frac{a^2-ab+b^2}{ab+4}~\text{and}~k=\{1,2,3,4\}$$ Simply trying them yields $k=4$, $a-b=2$ as the only solution. $\Rightarrow a+b=2020, a-b=2\rightarrow a=1011, b=1009$. And its symmetrical is a solution too. (We let $a\geq b$ at the beginning, now we are going back to it.)

$$101|a+b$$by legandre symbol similarity $$5|a+b$$

lemma 1 : let $p$ is a prime and $p|\frac{a^3+b^3}{a+b}, (a,b)=1$ then $p=3,3k+1$ it is very easy to check by orders ! lemma 2 : $(a,b)=2^\alpha$ let prime number $p \ne 2$ then $a^3+b^3=2020 (4+ab)$ but $p \nmid 4+ab$ so $p=5,101$ but $V_p(a^3+b^3) \ge 3$ and $V_p(2020) = 1$ and it isn't possible CASE 1: $a=2^\alpha.a'$ and $b=2^\beta.b'$ and $0 < \alpha < \beta$ then $V_2(2020(4+2^{\alpha+\beta})) = 4$ but $3 \mid V_2(a^3+b^3)$ and it isn't possible CASE 2: $a=2^\alpha.a'$ and $b=2^\beta.b'$ and $0 < \alpha = \beta$ $2^{3\alpha} (a'^3 + b'^3) = 2020(4+2^{2\alpha}a'b')$ and again as case 1 we can check $\alpha=1$ so $(a'+b')(a'^2+b'^2+a'b') = 505(1+a'b')$ and by lemma 1 we know $505 \nmid a'^2+b'^2+a'b' \Rightarrow a'^2+b'^2+a'b' \mid 1+a'b' \Rightarrow (a'-b')^2 \le 1 , a' \equiv b' (mod 2) \Rightarrow a'=b'$ and we can check this case dose not have any answers CASE 3:$(a,b)=1$ in this case again as case 2 wen can say : $2020 \nmid a'^2+b'^2+a'b' \Rightarrow a'^2+b'^2+a'b' \mid 1+a'b' \Rightarrow (a-b)^2 \le 2 $ and as case 2 : $a-b =2 \Rightarrow 2020 = a'+b' $ and the answer is $(a,b) = (1011,1009)$ note : $V_p(a) = ab$ means the biggest power of $p^b \mid a , p^{b+1} \nmid a$