Given a positive integer $ k$, there is a positive integer $ n$ with the property that one can obtain the sum of the first $ n$ positive integers by writing some $ k$ digits to the right of $ n$. Find the remainder of $ n$ when dividing at $ 9$.
1. We start by noting that the sum of the first $n$ positive integers is given by the formula:
   \[
   S = \frac{n(n+1)}{2}
   \]
2. According to the problem, there exists a positive integer $n$ such that appending $k$ digits to the right of $n$ results in the sum of the first $n$ positive integers. This can be expressed as:
   \[
   n \cdot 10^k + d = \frac{n(n+1)}{2}
   \]
   where $d$ is the $k$-digit number appended to $n$.

3. To find the remainder of $n$ when divided by $9$, we use the property of numbers modulo $9$. Specifically, the sum of the digits of a number is congruent to the number itself modulo $9$.

4. Consider the sum of the first $n$ positive integers modulo $9$:
   \[
   \frac{n(n+1)}{2} \equiv S \pmod{9}
   \]

5. We need to show that $n \cdot 10^k + d \equiv \frac{n(n+1)}{2} \pmod{9}$. Since $10 \equiv 1 \pmod{9}$, we have:
   \[
   10^k \equiv 1^k \equiv 1 \pmod{9}
   \]
   Therefore:
   \[
   n \cdot 10^k + d \equiv n + d \pmod{9}
   \]

6. Since $d$ is the $k$-digit number appended to $n$, it does not affect the sum modulo $9$ because $d$ is just a rearrangement of digits. Thus:
   \[
   n + d \equiv \frac{n(n+1)}{2} \pmod{9}
   \]

7. To find the remainder of $n$ when divided by $9$, we need to solve:
   \[
   n \equiv \frac{n(n+1)}{2} \pmod{9}
   \]

8. We test small values of $n$ to find a pattern:
   - For $n = 1$: $\frac{1(1+1)}{2} = 1 \equiv 1 \pmod{9}$
   - For $n = 2$: $\frac{2(2+1)}{2} = 3 \equiv 3 \pmod{9}$
   - For $n = 3$: $\frac{3(3+1)}{2} = 6 \equiv 6 \pmod{9}$
   - For $n = 4$: $\frac{4(4+1)}{2} = 10 \equiv 1 \pmod{9}$
   - For $n = 5$: $\frac{5(5+1)}{2} = 15 \equiv 6 \pmod{9}$
   - For $n = 6$: $\frac{6(6+1)}{2} = 21 \equiv 3 \pmod{9}$
   - For $n = 7$: $\frac{7(7+1)}{2} = 28 \equiv 1 \pmod{9}$
   - For $n = 8$: $\frac{8(8+1)}{2} = 36 \equiv 0 \pmod{9}$
   - For $n = 9$: $\frac{9(9+1)}{2} = 45 \equiv 0 \pmod{9}$

9. From the above, we observe that $n \equiv 1 \pmod{9}$ for $n = 1, 4, 7, \ldots$. Therefore, the remainder of $n$ when divided by $9$ is:
   \[
   \boxed{1}
   \]