Find the sum of all positive integers less than $2006$ which are both multiples of six and one more than a multiple of seven.
1. We need to find the sum of all positive integers less than $2006$ which are both multiples of six and one more than a multiple of seven. Let these integers be denoted by $x$.
2. We start with the condition that $x$ is a multiple of six, so $x = 6n$ for some integer $n$.
3. Additionally, $x$ is one more than a multiple of seven, so $x = 7m + 1$ for some integer $m$.
4. Equating the two expressions for $x$, we get:
   \[
   6n = 7m + 1
   \]
5. To solve for $n$ in terms of $m$, we take the equation modulo 7:
   \[
   6n \equiv 1 \pmod{7}
   \]
6. We need to find the multiplicative inverse of 6 modulo 7. The multiplicative inverse of 6 modulo 7 is the number $k$ such that:
   \[
   6k \equiv 1 \pmod{7}
   \]
   By trial, we find that $k = 6$ works because:
   \[
   6 \cdot 6 = 36 \equiv 1 \pmod{7}
   \]
7. Therefore, $n \equiv 6 \pmod{7}$, which means:
   \[
   n = 7k + 6 \quad \text{for some integer } k
   \]
8. Substituting $n = 7k + 6$ back into $x = 6n$, we get:
   \[
   x = 6(7k + 6) = 42k + 36
   \]
9. Thus, the numbers we are looking for are of the form $42k + 36$.
10. We need to find all such numbers less than $2006$. Solving for $k$:
    \[
    42k + 36 < 2006
    \]
    \[
    42k < 1970
    \]
    \[
    k < \frac{1970}{42} \approx 46.9048
    \]
    So, $k$ can take integer values from $0$ to $46$.
11. The sequence of numbers is $36, 78, 120, \ldots, 1968$. This is an arithmetic sequence with the first term $a = 36$ and common difference $d = 42$.
12. The number of terms in the sequence is $47$ (from $k = 0$ to $k = 46$).
13. The sum of an arithmetic sequence is given by:
    \[
    S_n = \frac{n}{2} (2a + (n-1)d)
    \]
    Substituting $n = 47$, $a = 36$, and $d = 42$:
    \[
    S_{47} = \frac{47}{2} (2 \cdot 36 + (47-1) \cdot 42)
    \]
    \[
    S_{47} = \frac{47}{2} (72 + 1932)
    \]
    \[
    S_{47} = \frac{47}{2} \cdot 2004
    \]
    \[
    S_{47} = 47 \cdot 1002
    \]
    \[
    S_{47} = 47094
    \]

The final answer is $\boxed{47094}$.