The function $g$, with domain and real numbers, fulfills the following:
$\bullet$ $g (x) \le x$, for all real $x$
$\bullet$ $g (x + y)  \le g (x) + g (y)$ for all  real $x,y$
Find $g (1990)$.
1. **Given Conditions:**
   - \( g(x) \le x \) for all real \( x \).
   - \( g(x + y) \le g(x) + g(y) \) for all real \( x, y \).

2. **Setting \( x = 0 \):**
   - From the first condition, \( g(0) \le 0 \).
   - From the second condition, \( g(0 + 0) \le g(0) + g(0) \), which simplifies to \( g(0) \le 2g(0) \). This implies \( g(0) \ge 0 \) because if \( g(0) \) were negative, \( 2g(0) \) would be more negative, contradicting \( g(0) \le 2g(0) \).

   Combining these, we get \( g(0) = 0 \).

3. **Analyzing \( g(x) \) and \( g(-x) \):**
   - Consider \( g(x) \) and \( g(-x) \):
     \[
     g(0) = g(x + (-x)) \le g(x) + g(-x) \implies 0 \le g(x) + g(-x)
     \]
   - From the first condition, \( g(x) \le x \) and \( g(-x) \le -x \). Adding these inequalities:
     \[
     g(x) + g(-x) \le x - x = 0
     \]
   - Combining the two results:
     \[
     0 \le g(x) + g(-x) \le 0 \implies g(x) + g(-x) = 0
     \]
   - Therefore, \( g(-x) = -g(x) \), showing that \( g \) is an odd function.

4. **Considering \( g(x) \le x \):**
   - We already have \( g(x) \le x \) from the given condition.
   - To show \( g(x) = x \), assume for contradiction that \( g(x) < x \) for some \( x \).

5. **Using the subadditivity condition:**
   - For any \( x \) and \( y \):
     \[
     g(x + y) \le g(x) + g(y)
     \]
   - Set \( y = x \):
     \[
     g(2x) \le g(x) + g(x) = 2g(x)
     \]
   - By induction, for any integer \( n \):
     \[
     g(nx) \le ng(x)
     \]
   - Since \( g(x) \le x \), we have:
     \[
     g(nx) \le nx
     \]

6. **Considering negative values:**
   - For negative \( x \), since \( g \) is odd:
     \[
     g(-x) = -g(x)
     \]
   - And since \( g(x) \le x \):
     \[
     -g(x) \le -x \implies g(-x) \le -x
     \]

7. **Conclusion:**
   - Since \( g(x) \le x \) and \( g(x) \ge x \) (from the subadditivity and odd function properties), we conclude:
     \[
     g(x) = x
     \]

Thus, \( g(1990) = 1990 \).

The final answer is \( \boxed{1990} \)