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 $\boxed{36018}$.