Areas of a triangle divided by 2 cevians into 4 regions, values of $A+B+C+D$ if $C=\frac{(2A+B+D)BD}{A^2-BD}$

Not a complete answer, but here is an $O(Nd(N^2))$ algorithm to find all solutions to $$(2A+B+C+D)BD=A^2C$$ with $A,B,C,D$ positive integers with sum $N$. Here $d(N^2)$ counts the number of divisors of $N^2$.

We rewrite the equation as $(A+N)BD=A^2C$. We get $A^2C\equiv0\mod A+N$, so since $A^2\equiv N^2\mod A+N$, we get $N^2C\equiv0\mod A+N$. It follows that $A+N=xy$ for some $x|N^2$ and $y|C$.

We thus iterate all triples $(x,y,z)$ of positive integers with $x|N^2$ and with $A,C\geq1$ and $A+C+2\leq N$ for $A=xy-N$ and $C=yz$. Note that the requirements $N+1\leq xy\leq 2N$ and $yz\leq N$ make sure that there are $O(N)$ possible integers $y,z$ for each possible $x$, so there are $O(Nd(N^2))$ possible triples.

We find $B+D=N-A-C=\beta$ and $BD=A^2C/(A+N)=\gamma$, so $(X-B)(X-D)=X^2-\beta X+\gamma$. For $\Delta=\beta^2-4\gamma$, we find that $B$ and $D$ are $(\beta+\sqrt\Delta)/2$ and $(\beta-\sqrt\Delta)/2$ in any order.

We thus simply need to check if $\Delta$ is a perfect square. If $\Delta=k^2$, note that $\beta$ and $k$ have the same parity, so since $\Delta<\beta^2$, it follows that $(B,D)=((\beta+k)/2,(\beta-k)/2)$ and vice versa always give valid solutions.

The following is an implementation in Python:

from math import isqrt

def solve(N):
    ds = [N]
    for d in range(1,N):
        if (N*N) % d == 0:
            ds += [d, N*N//d]
    solutions = []
    for x in ds:
        for y in range(max((N+x)//x,1), (2*N-2)//(x+1)+1):
            A = x*y-N
            for C in range(y, N-A-1, y):
                b, c = N-A-C, A*A*C//(A+N)
                d = b*b-4*c
                if d < 0:
                    continue
                k = isqrt(d)
                if k*k != d:
                    continue
                B, D = (b+k)//2, (b-k)//2
                solutions += [[A,B,C,D]]
                if k != 0:
                    solutions += [[A,D,C,B]]
    return solutions

For $N\leq5000$, there are $2753$ values of $N$ with a solution. It appears as though you got a very lucky streak with $2020,\ldots,2025$, because streaks of length $6$ are evidently quite rare. Looking through which values of $N$ do or do not have a solution seems to give me little insight, but anyone else is free to try and run this program and look for themselves. I will note that generally numbers with more divisors seem to have more solutions than numbers with few divisors, which makes sense, since there are more options for $x$.

In particular, let me prove for $N$ prime that a solution can not exist. We get $x\in\{1,N,N^2\}$. If $x=1$, then, since $y\leq N$, we have $A=xy-N\leq0$, a contradiction. If $x=N$ or $x=N^2$, then $A=xy-N$ is a multiple of $N$, a contradiction.

As a final note, we have $\sum_{n=1}^{N}d(n^2)=O(N\log^2N)$, so this gives an $O(N^2\log^2N)$ algorithm of finding all solutions with $A+B+C+D\leq N$. See Asymptotics of $\sum_{k=1}^nd(k^s)$, the sum of the divisor counts of the squares, cubes, etc.

EDIT: The following is a proof that any solution with $N=2p$ for $p$ prime must have $N=6$.

We get $x\in\{1,2,4,p,2p,4p,p^2,2p^2,4p^2\}$. Again, if $x=1$, then $A\leq0$, and if $N|x$, then $N|A$, so we only need to consider $x\in\{2,4,p,p^2\}$.

First, consider $x=p^2$. We get $A=(yp-2)p$, so we need $yp-2=1$, the only solution to which is $y=1$ and $p=3$, giving $N=6$.

Next, consider $x=p$. We get $A=(y-2)p$, so we need $y=3$. Then $\beta=p-3z$ and $\gamma=pz$, so $\Delta=p^2-10pz+9z^2=k^2$ for some integer $k\geq0$. We get $(p-10z)p=(k-3z)(k+3z)$. One of $k-3z$ and $k+3z$ must be divisible by $p$, and the product is $(p-10z)p<p^2$, so the other is less than $p$. Note that $\Delta=(p-9z)(p-z)\geq0$ gives $9z\leq p$, so $k-3z$ and $k+3z$ differ by less than $p$. It follows that $k+3z=p$ and $k-3z=p-10z$, which implies $z=0$, a contradiction.

Next, consider $x=2$. We get $A=2d$ for $d=y-p$, so we need $y>p$, and thus $z=1$. Then $\beta=p-3d$ and $\gamma=2d^2$, so $\Delta=p^2-6pd+d^2=k^2$ for some integer $k\geq0$. We get $(p-6d)p=(k-d)(k+d)$. One of $k-d$ and $k+d$ must be divisible by $p$, and the product is $(p-6d)p<p^2$, so the other is less than $p$. Note that $\Delta\geq0$ gives $(3+2\sqrt2)d\leq p$, so $k-d$ and $k+d$ differ by less than $p$. It follows that $k+d=p$ and $k-d=p-6d$, which implies $d=0$, a contradiction.

Finally, the most difficult case is $x=4$. We get $A=2(2y-p)$, so we need $y>p/2$, and thus $z<4$. Then $\beta=4p-(4+z)y$ and $\gamma=z(2y-p)^2$, so $$\Delta=(16-4z)p^2-(32-8z)py+(16-8z+z^2)y^2=k^2$$ for some integer $k\geq0$. We consider the three cases $z\in\{1,2,3\}$. We get $$z=1\rightarrow\Delta=12p^2-24py+9y^2=k^2,$$ $$z=2\rightarrow\Delta=8p^2-16py+4y^2=k^2,$$ $$z=3\rightarrow\Delta=4p^2-8py+y^2=k^2.$$

We start with the case $z=3$. We get $4p(p-8y)=(k-y)(k+y)$. One of $k-y$ and $k+y$ must be divisible by $p$. Note that $yz=C<N$ gives $y<2p/3$, so $k-y$ and $k+y$ differ by at most $4p/3<2p$. Furthermore, their product is $4p(p-8y)<0$, so $k+y$ is positive and $k-y$ is negative. It follows that either $k+y=p$ and $k-y=4(p-8y)$ or $k+y=4(8y-p)$ and $k-y=-p$. In either case, their difference is $2y=32y-3p$, so $3p=30y$, which is impossible, since $p$ is prime.

Next is the case $z=2$. Note that $k$ must be even, so we can assume $2p^2-4py+y^2=k^2$, giving $2p(p-2y)=(k-y)(k+y)$. One of $k-y$ and $k+y$ must be divisible by $p$. Note that $yz=C<N$ gives $y<p$, so $k-y$ and $k+y$ differ by less than $2p$. Furthermore, their product is $2p(p-y)<0$, so $k+y$ is positive and $k-y$ is negative. It follows that either $k+y=p$ and $k-y=2(p-2y)$ or $k+y=2(2y-p)$ and $k-y=-p$. In either case, their difference is $2y=4y-p$, so $p=2y$. This is only possible if $p=2$, but there are no solutions for $N=4$.

Finally, we consider the case $z=1$. We get $12p(p-2y)=(k-3y)(k+3y)$. One of $k-3y$ and $k+3y$ must be divisible by $p$. Note that $\Delta=3(2p-y)(2p-3y)\geq0$ gives $3y\leq2p$, so $k-3y$ and $k+3y$ differ by at most $4p$. Furthermore, their product is $12p(p-2y)<0$, so $k+3y$ is positive and $k-3y$ is negative. Whichever is a multiple of $p$ must thus be less than $4p$ in absolute value. We get three cases again:

If $k+3y=p$ and $k-3y=12(p-2y)$ or $k+3y=12(2y-p)$ and $k-3y=-p$, then their difference is $6y=24y-11p$, so $11p=18y$, a contradiction.

If $k+3y=2p$ and $k-3y=6(p-2y)$ or $k+3y=6(2y-p)$ and $k-3y=-2p$, then their difference is $6y=12y-4p$, so $4p=6y$, so $p=3$, giving $N=6$ again.

Finally, if $k+3y=3p$ and $k-3y=4(p-2y)$ or $k+3y=4(2y-p)$ and $k-3y=-3p$, then their difference is $6y=8y-p$, so $p=2y$. This is only possible if $p=2$, but there are no solutions for $N=4$.

Here's a quick-ish proof @SmileyCraft's result that the total area can be $2p$, for odd prime $p$, only if $p=3$. ("If" is satisfied by $(a,b,d)=(2,1,1)$.) In particular, the total area cannot be $2026=2\cdot 1013$.

First, as noted in my comment to the question, given positive-integer areas $a$, $b$, $d$, the total area is $$t := a+b+c+d = \frac{a(a+b)(a+d)}{a^2-bd} \tag1$$ So, if $t=2p$, then at least one of $a$, $a+b$, or $a+d$ must be divisible by $p$; moreover, since these quantities represent proper sub-areas of total area $2p$, "divisible by $p$" means "equal to $p$" (and at most one of $a$, $b$, $d$ can be "divisible by $p$"). Let's consider cases:

• $a=p$. Substituting this, and $t=2p$, into $(1)$ yields $$p(p-b-d) = 3 bd \quad\to\quad p\mid 3 \quad\to\quad p = 3 \tag2$$ We can deduce further that $(a,b,d)=(3,1,1)$.

• $a+b=p$ (likewise, $a+d=p$). Substituting $b=p-a$ and $t=2p$ into $(1)$ yields $$2 d p=a(a+d) \quad\to\quad p\mid a+d\quad\to\quad a+d=p \quad\to\quad 2p=3a \quad\to\quad p=3\tag3$$ We find in particular, that $(a,b,d)=(2,1,1)$.

Done! $\square$