Let $n$ be a positive integer such that $12n^2+12n+11$ is a $4$-digit number with all $4$ digits equal. Determine the value of $n$.
1. We start with the given expression \(12n^2 + 12n + 11\) and note that it must be a 4-digit number with all digits equal. The possible 4-digit numbers with all digits equal are \(1111, 2222, 3333, 4444, 5555, 6666, 7777, 8888, 9999\).

2. We rewrite the expression \(12n^2 + 12n + 11\) in a more convenient form:
   \[
   12n^2 + 12n + 11 = 12(n^2 + n + 1) - 1
   \]
   This implies that \(12(n^2 + n + 1) - 1\) must be one of the 4-digit numbers listed above.

3. Adding 1 to each of these numbers, we get:
   \[
   1112, 2223, 3334, 4445, 5556, 6667, 7778, 8889, 10000
   \]
   We need to find which of these numbers is a multiple of 12.

4. Checking each number:
   - \(1112 \div 12 = 92.6667\) (not an integer)
   - \(2223 \div 12 = 185.25\) (not an integer)
   - \(3334 \div 12 = 277.8333\) (not an integer)
   - \(4445 \div 12 = 370.4167\) (not an integer)
   - \(5556 \div 12 = 463\) (an integer)
   - \(6667 \div 12 = 555.5833\) (not an integer)
   - \(7778 \div 12 = 648.1667\) (not an integer)
   - \(8889 \div 12 = 740.75\) (not an integer)
   - \(10000 \div 12 = 833.3333\) (not an integer)

   The only number that is a multiple of 12 is \(5556\).

5. Setting \(12(n^2 + n + 1) = 5556\), we solve for \(n\):
   \[
   n^2 + n + 1 = \frac{5556}{12} = 463
   \]
   This gives us the quadratic equation:
   \[
   n^2 + n + 1 = 463
   \]
   \[
   n^2 + n - 462 = 0
   \]

6. Solving the quadratic equation using the quadratic formula \(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
   \[
   n = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-462)}}{2 \cdot 1}
   \]
   \[
   n = \frac{-1 \pm \sqrt{1 + 1848}}{2}
   \]
   \[
   n = \frac{-1 \pm \sqrt{1849}}{2}
   \]
   \[
   n = \frac{-1 \pm 43}{2}
   \]
   This gives us two solutions:
   \[
   n = \frac{42}{2} = 21 \quad \text{and} \quad n = \frac{-44}{2} = -22
   \]
   Since \(n\) must be a positive integer, we have \(n = 21\).

The final answer is \( \boxed{21} \).