Find all positive integers $m$ for which $2001\cdot S (m) = m$ where $S(m)$ denotes the sum of the digits of $m$.
To solve the problem, we need to find all positive integers \( m \) such that \( 2001 \cdot S(m) = m \), where \( S(m) \) denotes the sum of the digits of \( m \).

1. **Express \( m \) in terms of \( k \):**
   Let \( m = 2001k \). Then, the equation becomes:
   \[
   2001 \cdot S(2001k) = 2001k
   \]
   Dividing both sides by 2001, we get:
   \[
   S(2001k) = k
   \]

2. **Estimate the upper bound for \( k \):**
   The sum of the digits of \( 2001k \) is at most \( 9 \times (\text{number of digits in } 2001k) \). The number of digits in \( 2001k \) is approximately \( \lfloor \log_{10}(2001k) \rfloor + 1 \).

   Since \( 2001 \) has 4 digits, \( 2001k \) will have at most \( \lfloor \log_{10}(2001) + \log_{10}(k) \rfloor + 1 \) digits. For simplicity, we can approximate:
   \[
   \log_{10}(2001) \approx 3.301 \quad \text{and} \quad \log_{10}(k) \leq \log_{10}(50) \approx 1.699
   \]
   Thus, the number of digits in \( 2001k \) is at most:
   \[
   \lfloor 3.301 + 1.699 \rfloor + 1 = 5
   \]
   Therefore, the sum of the digits \( S(2001k) \) is at most:
   \[
   9 \times 5 = 45
   \]
   Hence, \( k \leq 45 \).

3. **Check values of \( k \):**
   We need to check values of \( k \) from 1 to 45 to see if \( S(2001k) = k \).

   - For \( k = 1 \):
     \[
     m = 2001 \times 1 = 2001 \quad \text{and} \quad S(2001) = 2 + 0 + 0 + 1 = 3 \neq 1
     \]
   - For \( k = 2 \):
     \[
     m = 2001 \times 2 = 4002 \quad \text{and} \quad S(4002) = 4 + 0 + 0 + 2 = 6 \neq 2
     \]
   - For \( k = 3 \):
     \[
     m = 2001 \times 3 = 6003 \quad \text{and} \quad S(6003) = 6 + 0 + 0 + 3 = 9 \neq 3
     \]
   - For \( k = 4 \):
     \[
     m = 2001 \times 4 = 8004 \quad \text{and} \quad S(8004) = 8 + 0 + 0 + 4 = 12 \neq 4
     \]
   - For \( k = 5 \):
     \[
     m = 2001 \times 5 = 10005 \quad \text{and} \quad S(10005) = 1 + 0 + 0 + 0 + 5 = 6 \neq 5
     \]
   - For \( k = 6 \):
     \[
     m = 2001 \times 6 = 12006 \quad \text{and} \quad S(12006) = 1 + 2 + 0 + 0 + 6 = 9 \neq 6
     \]
   - For \( k = 7 \):
     \[
     m = 2001 \times 7 = 14007 \quad \text{and} \quad S(14007) = 1 + 4 + 0 + 0 + 7 = 12 \neq 7
     \]
   - For \( k = 8 \):
     \[
     m = 2001 \times 8 = 16008 \quad \text{and} \quad S(16008) = 1 + 6 + 0 + 0 + 8 = 15 \neq 8
     \]
   - For \( k = 9 \):
     \[
     m = 2001 \times 9 = 18009 \quad \text{and} \quad S(18009) = 1 + 8 + 0 + 0 + 9 = 18 \neq 9
     \]
   - For \( k = 10 \):
     \[
     m = 2001 \times 10 = 20010 \quad \text{and} \quad S(20010) = 2 + 0 + 0 + 1 + 0 = 3 \neq 10
     \]
   - For \( k = 11 \):
     \[
     m = 2001 \times 11 = 22011 \quad \text{and} \quad S(22011) = 2 + 2 + 0 + 1 + 1 = 6 \neq 11
     \]
   - For \( k = 12 \):
     \[
     m = 2001 \times 12 = 24012 \quad \text{and} \quad S(24012) = 2 + 4 + 0 + 1 + 2 = 9 \neq 12
     \]
   - For \( k = 13 \):
     \[
     m = 2001 \times 13 = 26013 \quad \text{and} \quad S(26013) = 2 + 6 + 0 + 1 + 3 = 12 \neq 13
     \]
   - For \( k = 14 \):
     \[
     m = 2001 \times 14 = 28014 \quad \text{and} \quad S(28014) = 2 + 8 + 0 + 1 + 4 = 15 \neq 14
     \]
   - For \( k = 15 \):
     \[
     m = 2001 \times 15 = 30015 \quad \text{and} \quad S(30015) = 3 + 0 + 0 + 1 + 5 = 9 \neq 15
     \]
   - For \( k = 16 \):
     \[
     m = 2001 \times 16 = 32016 \quad \text{and} \quad S(32016) = 3 + 2 + 0 + 1 + 6 = 12 \neq 16
     \]
   - For \( k = 17 \):
     \[
     m = 2001 \times 17 = 34017 \quad \text{and} \quad S(34017) = 3 + 4 + 0 + 1 + 7 = 15 \neq 17
     \]
   - For \( k = 18 \):
     \[
     m = 2001 \times 18 = 36018 \quad \text{and} \quad S(36018) = 3 + 6 + 0 + 1 + 8 = 18 = 18
     \]

   Therefore, \( k = 18 \) is a solution.

4. **Verify if there are any other solutions:**
   We have checked all values of \( k \) from 1 to 45, and only \( k = 18 \) satisfies the condition \( S(2001k) = k \).

The final answer is \( \boxed{36018} \).