Find the sum of all integers $x$, $x \ge 3$, such that $201020112012_x$ (that is, $201020112012$ interpreted as a base $x$ number) is divisible by $x-1$
1. We start by interpreting the number \(201020112012_x\) in base \(x\). The number \(201020112012_x\) can be expanded as:
   \[
   2x^{11} + 0x^{10} + 1x^9 + 0x^8 + 2x^7 + 0x^6 + 1x^5 + 1x^4 + 2x^3 + 0x^2 + 1x + 2
   \]

2. We need to find the sum of all integers \(x\), \(x \ge 3\), such that \(201020112012_x\) is divisible by \(x-1\). 

3. Observe that for any base \(x\), \(x \equiv 1 \pmod{x-1}\). This implies that \(x^n \equiv 1 \pmod{x-1}\) for all \(n \in \mathbb{Z}^{+}\). 

4. Therefore, the number \(201020112012_x\) modulo \(x-1\) simplifies to:
   \[
   2 \cdot 1^{11} + 0 \cdot 1^{10} + 1 \cdot 1^9 + 0 \cdot 1^8 + 2 \cdot 1^7 + 0 \cdot 1^6 + 1 \cdot 1^5 + 1 \cdot 1^4 + 2 \cdot 1^3 + 0 \cdot 1^2 + 1 \cdot 1 + 2
   \]
   Simplifying this, we get:
   \[
   2 + 0 + 1 + 0 + 2 + 0 + 1 + 1 + 2 + 0 + 1 + 2 = 12
   \]

5. We need \(201020112012_x \equiv 0 \pmod{x-1}\). This means \(12 \equiv 0 \pmod{x-1}\), so \(x-1\) must be a divisor of 12.

6. The divisors of 12 are \(1, 2, 3, 4, 6, 12\). Therefore, the possible values of \(x\) are:
   \[
   x-1 = 1 \implies x = 2 \quad (\text{not valid since } x \ge 3)
   \]
   \[
   x-1 = 2 \implies x = 3
   \]
   \[
   x-1 = 3 \implies x = 4
   \]
   \[
   x-1 = 4 \implies x = 5
   \]
   \[
   x-1 = 6 \implies x = 7
   \]
   \[
   x-1 = 12 \implies x = 13
   \]

7. The valid values of \(x\) are \(3, 4, 5, 7, 13\).

8. Summing these values, we get:
   \[
   3 + 4 + 5 + 7 + 13 = 32
   \]

The final answer is \(\boxed{32}\).