Assume to the contrary that there exists an integer $A = \overline{a_m \dots a_1}$ such that $\overline{a_m \dots a_1}$ has at most 2 different digits and $d \mid A$, where $d$ denotes $10^n - 3$. Consider $S = \overline{a_n \dots a_1} + 3 \times \overline{a_m \dots a_{n+1}}$ which exists since $m > n$. It follows from $d \mid A$ that $d \mid S$. Moreover, the assumption that $2n-2 \ge m$ yields $S \le 10^n - 1 + 3 \times (10^{n-2} - 1) < 2d$ and so we must have
$$\overline{a_n \dots a_1} + 3 \times \overline{a_m \dots a_{n+1}} = 10^n - 3. \quad (0.1)$$
Hence $a_1 + 3a_{n+1} \equiv 7 \pmod{10}$ and consequently, $a_1 \neq a_{n+1}$. Observe that
$$\overline{a_n \dots a_1} = d - 3 \times \overline{a_m \dots a_{n+1}} > d - 3 \times (10^{n-2} - 1) > 9 \times 10^{n-1},$$
which implies $a_n = 9$. Thus $9 \in \{a_1, a_{n+1}\}$ by the assumption on $A$. If $a_{n+1} = 9$ then $a_1 = 0$ and so $A$ is divisible by 9, contradicting to (0.1). If $a_1 = 9$ then $a_{n+1} = 6$, giving a similar contradiction.