Find the least positive integer such that the sum of its digits is $2011$ and the product of its digits is a power of $6$.
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 \(\boxed{3488888\ldots8889999\ldots9999}\).