We will find a number $n$ in the form $10^k(10a + b) + c$ where $0 \le c < 10^k$ and $a, b$ are digits. Deleting the digit $b$ gives the number $n_1 = 10^k a + c$. Since $n - n_1 = 10^k(9a+b)$, to satisfy the conditions of the problem it is sufficient that $d$ divides $9a+b$ and $10^k a + c$.

The digit $b$ can be chosen so that $9$ divides $d-b$ and we set $a = \frac{d-b}{9}$. Then $d = 9a+b$. Let $k$ be such that $10^{k-1} > d$. Then there are integers $q$ and $0 \le r < d$ such that $10^k a + 10^{k-1} = dq + r$. If we set $c = 10^{k-1} - r > 0$, then $10^k a + c = dq$, obviously a number divisible by $d$.