Count the number $N$ of all sets $A:=\{x_1,x_2,x_3,x_4\}$ of non-negative integers satisfying
$$x_1+x_2+x_3+x_4=36$$
in at least four different ways.

[i]Proposed by Eugene J. Ionaşcu[/i]
To solve the problem, we need to count the number of sets \( A = \{x_1, x_2, x_3, x_4\} \) of non-negative integers such that \( x_1 + x_2 + x_3 + x_4 = 36 \). This is a classic problem that can be solved using the stars and bars method or generating functions. Here, we will use generating functions.

1. **Generating Function Setup:**
   The generating function for each \( x_i \) (where \( x_i \) is a non-negative integer) is given by:
   \[
   \frac{1}{1 - x}
   \]
   Since we have four variables \( x_1, x_2, x_3, x_4 \), the generating function for the sum \( x_1 + x_2 + x_3 + x_4 \) is:
   \[
   \left( \frac{1}{1 - x} \right)^4 = \frac{1}{(1 - x)^4}
   \]

2. **Finding the Coefficient:**
   We need to find the coefficient of \( x^{36} \) in the expansion of \( \frac{1}{(1 - x)^4} \). The general form of the generating function \( \frac{1}{(1 - x)^k} \) is:
   \[
   \frac{1}{(1 - x)^k} = \sum_{n=0}^{\infty} \binom{n + k - 1}{k - 1} x^n
   \]
   For our case, \( k = 4 \), so:
   \[
   \frac{1}{(1 - x)^4} = \sum_{n=0}^{\infty} \binom{n + 3}{3} x^n
   \]

3. **Extracting the Coefficient:**
   We need the coefficient of \( x^{36} \) in the series expansion:
   \[
   \sum_{n=0}^{\infty} \binom{n + 3}{3} x^n
   \]
   Therefore, the coefficient of \( x^{36} \) is:
   \[
   \binom{36 + 3}{3} = \binom{39}{3}
   \]

4. **Calculating the Binomial Coefficient:**
   We calculate \( \binom{39}{3} \):
   \[
   \binom{39}{3} = \frac{39 \times 38 \times 37}{3 \times 2 \times 1} = \frac{54834}{6} = 9139
   \]

Conclusion:
The number of sets \( A = \{x_1, x_2, x_3, x_4\} \) of non-negative integers such that \( x_1 + x_2 + x_3 + x_4 = 36 \) is \( 9139 \).

The final answer is \( \boxed{9139} \).