Find the largest natural number $n$ such that for all real numbers $a, b, c, d$ the following holds:
$$(n + 2)\sqrt{a^2 + b^2} + (n + 1)\sqrt{a^2 + c^2} + (n + 1)\sqrt{a^2 + d^2} \ge  n(a + b + c + d)$$
1. **Understanding the problem**: We need to find the largest natural number \( n \) such that the inequality 
   \[
   (n + 2)\sqrt{a^2 + b^2} + (n + 1)\sqrt{a^2 + c^2} + (n + 1)\sqrt{a^2 + d^2} \ge n(a + b + c + d)
   \]
   holds for all real numbers \( a, b, c, d \).

2. **Using the Cauchy-Schwarz Inequality**: To approach this problem, we can use the Cauchy-Schwarz inequality in the form:
   \[
   \left( \sum_{i=1}^n x_i y_i \right)^2 \leq \left( \sum_{i=1}^n x_i^2 \right) \left( \sum_{i=1}^n y_i^2 \right)
   \]
   Let's apply this to the terms involving square roots.

3. **Applying Cauchy-Schwarz**: Consider the terms \(\sqrt{a^2 + b^2}\), \(\sqrt{a^2 + c^2}\), and \(\sqrt{a^2 + d^2}\). We can write:
   \[
   \sqrt{a^2 + b^2} \ge \frac{|a| + |b|}{\sqrt{2}}
   \]
   \[
   \sqrt{a^2 + c^2} \ge \frac{|a| + |c|}{\sqrt{2}}
   \]
   \[
   \sqrt{a^2 + d^2} \ge \frac{|a| + |d|}{\sqrt{2}}
   \]

4. **Combining the inequalities**: Substitute these into the original inequality:
   \[
   (n + 2)\frac{|a| + |b|}{\sqrt{2}} + (n + 1)\frac{|a| + |c|}{\sqrt{2}} + (n + 1)\frac{|a| + |d|}{\sqrt{2}} \ge n(a + b + c + d)
   \]

5. **Simplifying the inequality**: Combine the terms involving \( |a| \):
   \[
   \frac{(n + 2)|a| + (n + 1)|a| + (n + 1)|a| + (n + 2)|b| + (n + 1)|c| + (n + 1)|d|}{\sqrt{2}} \ge n(a + b + c + d)
   \]
   \[
   \frac{(3n + 4)|a| + (n + 2)|b| + (n + 1)|c| + (n + 1)|d|}{\sqrt{2}} \ge n(a + b + c + d)
   \]

6. **Analyzing the coefficients**: For the inequality to hold for all \( a, b, c, d \), the coefficients on the left must be at least as large as those on the right. This gives us:
   \[
   \frac{3n + 4}{\sqrt{2}} \ge n, \quad \frac{n + 2}{\sqrt{2}} \ge n, \quad \frac{n + 1}{\sqrt{2}} \ge n
   \]

7. **Solving for \( n \)**: Solve each inequality separately:
   \[
   \frac{3n + 4}{\sqrt{2}} \ge n \implies 3n + 4 \ge n\sqrt{2} \implies 3n - n\sqrt{2} \ge -4 \implies n(3 - \sqrt{2}) \ge -4
   \]
   \[
   \frac{n + 2}{\sqrt{2}} \ge n \implies n + 2 \ge n\sqrt{2} \implies n(1 - \sqrt{2}) \ge -2
   \]
   \[
   \frac{n + 1}{\sqrt{2}} \ge n \implies n + 1 \ge n\sqrt{2} \implies n(1 - \sqrt{2}) \ge -1
   \]

8. **Finding the largest \( n \)**: The most restrictive inequality is \( n(1 - \sqrt{2}) \ge -1 \). Solving for \( n \):
   \[
   n \le \frac{1}{\sqrt{2} - 1} = \frac{1}{\sqrt{2} - 1} \cdot \frac{\sqrt{2} + 1}{\sqrt{2} + 1} = \frac{\sqrt{2} + 1}{1} = \sqrt{2} + 1 \approx 2.414
   \]
   Since \( n \) must be a natural number, the largest possible \( n \) is \( 2 \).

The final answer is \( \boxed{2} \).