Find all possible integer values of $n$ such that $12n^2 + 12n + 11$ is a $4$-digit number with equal digits.
1. We start with the given equation \(12n^2 + 12n + 11\) and we need it to be a 4-digit number with equal digits. The possible 4-digit numbers with equal digits are \(1111, 2222, 3333, 4444, 5555, 6666, 7777, 8888, 9999\).

2. Let \(12n^2 + 12n + 11 = 1111k\), where \(k\) is an integer from 1 to 9. We need to find \(k\) such that \(12n^2 + 12n + 11\) is a multiple of 1111.

3. First, we take the equation modulo 4:
   \[
   12n^2 + 12n + 11 \equiv 0 \pmod{4}
   \]
   Since \(12 \equiv 0 \pmod{4}\), we have:
   \[
   11 \equiv 0 \pmod{4} \implies 11 \equiv 3 \pmod{4}
   \]
   Therefore, \(k \equiv 1 \pmod{4}\).

4. Next, we take the equation modulo 3:
   \[
   12n^2 + 12n + 11 \equiv 0 \pmod{3}
   \]
   Since \(12 \equiv 0 \pmod{3}\), we have:
   \[
   11 \equiv 0 \pmod{3} \implies 11 \equiv 2 \pmod{3}
   \]
   Therefore, \(k \equiv 2 \pmod{3}\).

5. Combining the two congruences, we use the Chinese Remainder Theorem:
   \[
   k \equiv 1 \pmod{4}
   \]
   \[
   k \equiv 2 \pmod{3}
   \]
   Solving these, we find \(k = 5\).

6. Now, we substitute \(k = 5\) back into the equation:
   \[
   12n^2 + 12n + 11 = 1111 \times 5
   \]
   \[
   12n^2 + 12n + 11 = 5555
   \]
   \[
   12n^2 + 12n - 5544 = 0
   \]

7. We solve the quadratic equation \(12n^2 + 12n - 5544 = 0\):
   \[
   n^2 + n - 462 = 0
   \]
   Using the quadratic formula \(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 1\), and \(c = -462\):
   \[
   n = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-462)}}{2 \cdot 1}
   \]
   \[
   n = \frac{-1 \pm \sqrt{1 + 1848}}{2}
   \]
   \[
   n = \frac{-1 \pm \sqrt{1849}}{2}
   \]
   \[
   n = \frac{-1 \pm 43}{2}
   \]
   Therefore, the solutions are:
   \[
   n = \frac{42}{2} = 21
   \]
   \[
   n = \frac{-44}{2} = -22
   \]

The final answer is \(\boxed{n = 21, -22}\).