Solution:
Let $A=1000 a+100 b+10 c+d$. Then we obtain the equality
$$\frac{1}{3}(1000 a+100 b+10 c+d)+2000=\frac{2}{3}(1000 d+100 c+10 b+a)$$
Multiply both sides by $3$ to clear denominators:
$$1000 a+100 b+10 c+d + 6000 = 2000 d + 200 c + 20 b + 2 a$$
Bring all terms to one side:
$$1000 a + 100 b + 10 c + d + 6000 - 2000 d - 200 c - 20 b - 2 a = 0$$
Group like terms:
$$(1000 a - 2 a) + (100 b - 20 b) + (10 c - 200 c) + (d - 2000 d) + 6000 = 0$$
$$998 a + 80 b - 190 c - 1999 d + 6000 = 0$$
$$998 a + 80 b - 190 c + 6000 = 1999 d$$
So
$$1999 d + 190 c = 80 b + 998 a + 6000$$
It is clear that $d$ is an even digit and $d>2$. So we have to investigate three cases:

(i) $d=4$:
Comparing the last digits in the upper equality we see that $a=2$ or $a=1$.
If $a=2$ then $19 c=80$, which is possible only when $c=0$. Hence the number $A=2004$ satisfies the condition.
If $a=7$ then $19 c-8 b=490$, which is impossible.

(ii) $d=6$:
Then $190 c+5994=80 b+998 a$. Comparing the last digits we obtain that $a=3$ or $a=8$.
If $a=3$ then $80 b+998 a<80 \cdot 9+1000 \cdot 3<5994$.
If $a=8$ then $306+998 a \geq 998 \cdot 8=7984=5994+1990>5994+190 c$.

(iii) $d=8$:
Then $190 c+9992=80 b+998 a$. Now $80 b+998 a \leq 80 \cdot 9+998 \cdot 9=9702<9992+190 c$.

Hence we have the only solution $A=2004$.