A finite sequence of positive integers $ m_i$ for $ i\equal{}1,2,...,2006$ are defined so that $ m_1\equal{}1$ and $ m_i\equal{}10m_{i\minus{}1} \plus{}1$ for $ i>1$. How many of these integers are divisible by $ 37$?
1. **Define the sequence**: The sequence \( m_i \) is defined such that \( m_1 = 1 \) and \( m_i = 10m_{i-1} + 1 \) for \( i > 1 \).

2. **Express the sequence in a closed form**: We can observe that the sequence can be written as:
   \[
   m_1 = 1
   \]
   \[
   m_2 = 10 \cdot 1 + 1 = 11
   \]
   \[
   m_3 = 10 \cdot 11 + 1 = 111
   \]
   \[
   m_4 = 10 \cdot 111 + 1 = 1111
   \]
   and so on. This suggests that \( m_i \) can be written as:
   \[
   m_i = \sum_{k=0}^{i-1} 10^k = \frac{10^i - 1}{9}
   \]

3. **Determine divisibility by 37**: We need to find how many of these integers \( m_i \) are divisible by 37. This means we need to solve:
   \[
   \frac{10^i - 1}{9} \equiv 0 \pmod{37}
   \]
   which simplifies to:
   \[
   10^i - 1 \equiv 0 \pmod{37}
   \]
   or:
   \[
   10^i \equiv 1 \pmod{37}
   \]

4. **Find the order of 10 modulo 37**: The order of 10 modulo 37 is the smallest positive integer \( d \) such that \( 10^d \equiv 1 \pmod{37} \). By checking powers of 10 modulo 37, we find:
   \[
   10^1 \equiv 10 \pmod{37}
   \]
   \[
   10^2 \equiv 100 \equiv 26 \pmod{37}
   \]
   \[
   10^3 \equiv 260 \equiv 1 \pmod{37}
   \]
   Therefore, the order of 10 modulo 37 is 3.

5. **Count the multiples of 3 up to 2006**: Since \( 10^i \equiv 1 \pmod{37} \) when \( i \) is a multiple of 3, we need to count how many multiples of 3 are there in the range from 1 to 2006. This is given by:
   \[
   \left\lfloor \frac{2006}{3} \right\rfloor = 668
   \]

Conclusion:
\[
\boxed{668}
\]