Answer: 4050
Put $n = 2025$ and $d = 10^n + 3$. Let $N = \overline{a_m \dots a_2 a_1}$ be a multiple of $d$ such that the number of different digits in $N$ is at most two. Then we can assume that $a_1 \ne 0$ and clearly $m \ge n+1$ since $N \ge d$. Suppose that $2n > m$, and consider $A = \overline{a_n \dots a_1}$ and $B = \overline{a_m \dots a_{n+1}}$. Since $d > 3 \times 10^{n-1} > 3B$ and $d > A$ we have $|A - 3B| < d$. Moreover, $A - 3B = N - dB$ is a multiple of $d$ and so $A = 3B$, or equivalently,
$$\overline{a_n \dots a_1} = 3 \times \overline{a_m \dots a_{n+1}}. \qquad (0.2)$$
If $a_n = 0$ then $a_{n+1} = a_1$ since $a_{n+1} \ne 0$ and $a_1 \ne 0$. Therefore, $a_1 = 5$ and $3a_{n+2} + 1 \equiv a_2 \pmod{10}$ by (0.2). But this is impossible since $a_2, a_{n+2} \in \{0, 5\}$. Thus $a_n \ne 0$ and so it follows from (0.2) that $m = 2n - 1$. Moreover, we have $\overline{a_n a_{n-1}} = 3a_m + c$ for some digit $c$. Observe that $1 \le a_n \le 2$ and $0 \le c \le 2$ since $3B < 3 \times 10^{n-1}$ and $3 \times \overline{a_{m-1} \dots a_{n+1}} < 3 \times 10^{n-2}$.
Assume that $a_n = 2$. If $a_1 = 2$ then $a_{n+1} = 4$ by (0.2). Hence $c = \overline{a_n a_{n-1}} - 3a_m \ge 22 - 3 \times 4 = 10$, a contradiction. If $a_{n+1} = 2$ then $a_1 = 6$ and so $c \ge 22 - 3 \times 6 = 4$, a contradiction.
Assume that $a_n = 1$. If $a_{n+1} = 1$ then $a_1 = 3$, hence (0.2) is impossible. If $a_1 = 1$ then $a_{n+1} = 7$, hence $c \ge 11 - 3 \times 1 = 8$, a contradiction.
Thus $m \ge 2n$ and so it suffices to show that $2n$-digit number $\overline{1 \dots 11 \ 3 \dots 33}$ is divisible by $d$:
$$\begin{aligned} \overline{1 \dots 11 \ 3 \dots 33} &= \overline{1 \dots 11} \times 10^n + \overline{3 \dots 33} \\ &= \overline{1 \dots 11} \times (d-3) + 3 \times \overline{1 \dots 11} \equiv 0 \pmod{d}. \end{aligned}$$