Find all positive integers $n$ for which $$3x^n + n(x + 2) - 3 \geq  nx^2$$ holds for all real numbers $x.$
To find all positive integers \( n \) for which the inequality 
\[ 3x^n + n(x + 2) - 3 \geq nx^2 \]
holds for all real numbers \( x \), we will analyze the inequality step by step.

1. **Check for \( x = -1 \):**
   \[ 3(-1)^n + n(-1 + 2) - 3 \geq n(-1)^2 \]
   Simplifying, we get:
   \[ 3(-1)^n + n - 3 \geq n \]
   \[ 3(-1)^n - 3 \geq 0 \]
   This implies:
   \[ 3(-1)^n \geq 3 \]
   Since \( 3(-1)^n \) is 3 if \( n \) is even and -3 if \( n \) is odd, the inequality holds only if \( n \) is even. Therefore, \( n \) must be even.

2. **Assume \( n \) is even and \( x \geq 0 \):**
   For \( x = 0 \):
   \[ 3 \cdot 0^n + n(0 + 2) - 3 \geq n \cdot 0^2 \]
   \[ 0 + 2n - 3 \geq 0 \]
   \[ 2n - 3 \geq 0 \]
   \[ n \geq \frac{3}{2} \]
   Since \( n \) is a positive integer, \( n \geq 2 \).

   For \( x = 1 \):
   \[ 3 \cdot 1^n + n(1 + 2) - 3 \geq n \cdot 1^2 \]
   \[ 3 + 3n - 3 \geq n \]
   \[ 3n \geq n \]
   This is always true for any \( n \).

3. **Assume \( x \geq 2 \) and \( n = 2 \):**
   \[ 3x^2 + 2(x + 2) - 3 \geq 2x^2 \]
   \[ 3x^2 + 2x + 4 - 3 \geq 2x^2 \]
   \[ x^2 + 2x + 1 \geq 0 \]
   \[ (x + 1)^2 \geq 0 \]
   This is always true for any \( x \).

4. **Assume \( x \geq 2 \) and \( n \geq 4 \):**
   \[ 3x^n + n(x + 2) - 3 \geq nx^2 \]
   \[ 3x^n - nx^2 + nx + 2n - 3 \geq 0 \]
   It suffices to show:
   \[ 3x^n \geq nx^2 \]
   For \( x \geq 2 \) and \( n \geq 4 \), we can use induction to prove this.

   **Base case:** For \( n = 4 \):
   \[ 3x^4 \geq 4x^2 \]
   \[ 3x^2 \geq 4 \]
   \[ x^2 \geq \frac{4}{3} \]
   Since \( x \geq 2 \), \( x^2 \geq 4 \), which is greater than \(\frac{4}{3}\).

   **Inductive step:** Assume \( 3x^k \geq kx^2 \) for some \( k \geq 4 \). We need to show \( 3x^{k+1} \geq (k+1)x^2 \):
   \[ 3x^{k+1} \geq 3x \cdot kx^2 / 3 \]
   \[ 3x^{k+1} \geq kx^2 \cdot x \]
   \[ 3x^{k+1} \geq kx^3 \]
   Since \( x \geq 2 \), \( 3x \geq k \) for \( k \geq 4 \).

5. **Assume \( x \leq -2 \):**
   \[ 3x^n + n(x + 2) - 3 \geq nx^2 \]
   \[ 3x^n - nx^2 + nx + 2n - 3 \geq 0 \]
   It suffices to show:
   \[ 3x^n \geq nx^2 \]
   Substitute \( y = -x \):
   \[ 3(-y)^n \geq n(y^2) \]
   Since \( n \) is even, \( (-y)^n = y^n \):
   \[ 3y^n \geq ny^2 \]
   For \( y \geq 2 \) and \( n \geq 4 \), we can use induction to prove this.

   **Base case:** For \( n = 4 \):
   \[ 3y^4 \geq 4y^2 \]
   \[ 3y^2 \geq 4 \]
   \[ y^2 \geq \frac{4}{3} \]
   Since \( y \geq 2 \), \( y^2 \geq 4 \), which is greater than \(\frac{4}{3}\).

   **Inductive step:** Assume \( 3y^k \geq ky^2 \) for some \( k \geq 4 \). We need to show \( 3y^{k+1} \geq (k+1)y^2 \):
   \[ 3y^{k+1} \geq 3y \cdot ky^2 / 3 \]
   \[ 3y^{k+1} \geq ky^2 \cdot y \]
   \[ 3y^{k+1} \geq ky^3 \]
   Since \( y \geq 2 \), \( 3y \geq k \) for \( k \geq 4 \).

Therefore, the inequality holds for all even positive integers \( n \).

The final answer is \( \boxed{ n } \) is any even positive integer.