Let $k$ be a positive integer such that $1 +\frac12+\frac13+ ... +\frac{1}{13}=\frac{k}{13!}$. Find the remainder when $k$ is divided by $7$.
1. We start with the given equation:
   \[
   1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{13} = \frac{k}{13!}
   \]
   To find \( k \), we multiply both sides by \( 13! \):
   \[
   13! \left(1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{13}\right) = k
   \]
   This can be rewritten as:
   \[
   k = 13! + \frac{13!}{2} + \frac{13!}{3} + \cdots + \frac{13!}{13}
   \]

2. We need to find the remainder when \( k \) is divided by 7. Therefore, we compute:
   \[
   k \pmod{7}
   \]
   This is equivalent to:
   \[
   13! + \frac{13!}{2} + \frac{13!}{3} + \cdots + \frac{13!}{13} \pmod{7}
   \]

3. First, we simplify \( 13! \pmod{7} \). Note that:
   \[
   13! = 1 \cdot 2 \cdot 3 \cdot \cdots \cdot 13
   \]
   We can reduce each term modulo 7:
   \[
   13! \equiv 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 0 \cdot 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \pmod{7}
   \]
   Since \( 7 \equiv 0 \pmod{7} \), any product involving 7 or its multiples will be 0 modulo 7. Therefore:
   \[
   13! \equiv 0 \pmod{7}
   \]

4. Next, we consider the terms \( \frac{13!}{2}, \frac{13!}{3}, \ldots, \frac{13!}{13} \) modulo 7. Since \( 13! \equiv 0 \pmod{7} \), each of these terms will also be 0 modulo 7:
   \[
   \frac{13!}{2} \equiv 0 \pmod{7}, \quad \frac{13!}{3} \equiv 0 \pmod{7}, \quad \ldots, \quad \frac{13!}{13} \equiv 0 \pmod{7}
   \]

5. Summing these results, we get:
   \[
   k \equiv 0 + 0 + 0 + \cdots + 0 \pmod{7}
   \]
   Therefore:
   \[
   k \equiv 0 \pmod{7}
   \]

The final answer is \(\boxed{0}\)