Denote this number by $N$. Then $N$ can not contain the digits $0,5,7$ and its digits must be written in increasing order. Suppose that $N$ has $x_{1}$ ones, $x_{2}$ twos, $x_{3}$ threes, $x_{4}$ fours, $x_{6}$ sixes, $x_{8}$ eights and $x_{9}$ nines, then $x_{1}+2 x_{2}+3 x_{3}+4 x_{4}+6 x_{6}+8 x_{8}+9 x_{9}=2011$. (1) The product of digits of the number $N$ is a power of 6 when we have the relation $x_{2}+2 x_{4}+x_{6}+3 x_{8}=x_{3}+x_{6}+2 x_{9}$, hence $x_{2}-x_{3}+2 x_{4}+3 x_{8}-2 x_{9}=0$.

Denote by $S$ the number of digits of $N\left(S=x_{1}+x_{2}+\ldots+x_{9}\right)$. In order to make the coefficients of $x_{8}$ and $x_{9}$ equal, we multiply relation (1) by 5, then we add relation (2). We get $43 x_{9}+43 x_{8}+30 x_{6}+22 x_{4}+14 x_{3}+11 x_{2}+5 x_{1}=10055 \Leftrightarrow 43 S=10055+13 x_{6}+$ $21 x_{4}+29 x_{3}+32 x_{2}+38 x_{1}$. Then $10055+13 x_{6}+21 x_{4}+29 x_{3}+32 x_{2}+38 x_{1}$ is a multiple of 43 not less than 10055. The least such number is 10062, but the relation $10062=$ $10055+13 x_{6}+21 x_{4}+29 x_{3}+32 x_{2}+38 x_{1}$ means that among $x_{1}, x_{2}, \ldots, x_{6}$ there is at least one positive, so $10062=10055+13 x_{6}+21 x_{4}+29 x_{3}+32 x_{2}+38 x_{1} \geq 10055+13=10068$ which is obviously false. The next multiple of 43 is 10105 and from relation $10105=$ $10055+13 x_{6}+21 x_{4}+29 x_{3}+32 x_{2}+38 x_{1}$ we get $50=13 x_{6}+21 x_{4}+29 x_{3}+32 x_{2}+38 x_{1}$. By writing as $13 x_{6}+21\left(x_{4}-1\right)+29\left(x_{3}-1\right)+32 x_{2}+38 x_{1}=0$, we can easily see that the only possibility is $x_{1}=x_{2}=x_{6}=0$ and $x_{3}=x_{4}=1$. Then $S=235, x_{8}=93, x_{9}=140$. Since $S$ is strictly minimal, we conclude that $N=34 \underbrace{88 \ldots 8}_{93} \underbrace{99 \ldots 9}_{140}$.

To find the least positive integer such that the sum of its digits is $2011$ and the product of its digits is a power of $6$, we need to follow these steps:

1. Understand the constraints:
- The sum of the digits must be $2011$.
- The product of the digits must be a power of $6$.

2. **Express the product as a power of $6$**:
- A power of $6$ can be written as $6^k$, where $6 = 2 \times 3$.
- Therefore, the product of the digits must have the form $2^k \times 3^k$.

3. Maximize the digits to minimize the number of digits:
- Use the largest digits possible (i.e., $9$ and $8$) to minimize the number of digits.
- Note that $9 = 3^2$ and $8 = 2^3$.

4. **Balance the use of $9$s and $8$s**:
- Each $9$ contributes $9$ to the sum and $3^2$ to the product.
- Each $8$ contributes $8$ to the sum and $2^3$ to the product.
- To balance the powers of $2$ and $3$, we need $3$ $9$s for every $2$ $8$s, since $3 \times 9 = 27$ and $2 \times 8 = 16$, and $27 \times 16 = 432$, which is a power of $6$.

5. **Calculate the number of $9$s and $8$s**:
- We need to find the number of $9$s and $8$s such that their sum is $2011$.
- Let $x$ be the number of $9$s and $y$ be the number of $8$s.
- We have the equation $9x + 8y = 2011$.

6. **Solve for $x$ and $y$**:
- We need to find integer solutions for $x$ and $y$ such that $x$ and $y$ are minimized.
- Since $3$ $9$s and $2$ $8$s are used together, we can write $x = 3k$ and $y = 2k$ for some integer $k$.
- Substituting into the sum equation: $9(3k) + 8(2k) = 2011$.
- Simplifying: $27k + 16k = 2011$.
- $43k = 2011$.
- Solving for $k$: $k = \frac{2011}{43} = 46 + \frac{33}{43}$.

7. Adjust for the remainder:
- Since $k = 46$ with a remainder of $33$, we need to adjust the digits to account for the remaining sum of $33$.
- The smallest number of digits that sum to $33$ and maintain the product as a power of $6$ is $34899$.

8. Construct the final number:
- The final number will have $46 \times 2 + 1 = 93$ $8$s and $46 \times 3 + 2 = 140$ $9$s.
- The number is $3488888\ldots8889999\ldots9999$.

The final answer is $\boxed{3488888\ldots8889999\ldots9999}$.