Let $ f$ be a function defined on the set of non-negative integers and taking values in the same
set. Given that

(a) $ \displaystyle x \minus{} f(x) \equal{} 19\left[\frac{x}{19}\right] \minus{} 90\left[\frac{f(x)}{90}\right]$ for all non-negative integers $ x$;

(b) $ 1900 < f(1990) < 2000$,

find the possible values that $ f(1990)$ can take.
(Notation : here $ [z]$ refers to largest integer that is $ \leq z$, e.g. $ [3.1415] \equal{} 3$).
1. Given the function \( f \) defined on the set of non-negative integers, we start with the equation:
   \[
   x - f(x) = 19 \left\lfloor \frac{x}{19} \right\rfloor - 90 \left\lfloor \frac{f(x)}{90} \right\rfloor
   \]
   for all non-negative integers \( x \).

2. We need to find the possible values of \( f(1990) \) given that \( 1900 < f(1990) < 2000 \).

3. Plugging \( x = 1990 \) into the equation, we get:
   \[
   1990 - f(1990) = 19 \left\lfloor \frac{1990}{19} \right\rfloor - 90 \left\lfloor \frac{f(1990)}{90} \right\rfloor
   \]

4. Calculate \( \left\lfloor \frac{1990}{19} \right\rfloor \):
   \[
   \frac{1990}{19} \approx 104.7368 \implies \left\lfloor \frac{1990}{19} \right\rfloor = 104
   \]
   Therefore:
   \[
   1990 - f(1990) = 19 \cdot 104 - 90 \left\lfloor \frac{f(1990)}{90} \right\rfloor
   \]
   Simplifying:
   \[
   1990 - f(1990) = 1976 - 90 \left\lfloor \frac{f(1990)}{90} \right\rfloor
   \]

5. Rearrange to solve for \( f(1990) \):
   \[
   f(1990) = 1990 - 1976 + 90 \left\lfloor \frac{f(1990)}{90} \right\rfloor
   \]
   \[
   f(1990) = 14 + 90 \left\lfloor \frac{f(1990)}{90} \right\rfloor
   \]

6. Given \( 1900 < f(1990) < 2000 \), we divide by 90:
   \[
   \frac{1900}{90} < \frac{f(1990)}{90} < \frac{2000}{90}
   \]
   \[
   21.1111 < \frac{f(1990)}{90} < 22.2222
   \]

7. Therefore, \( \left\lfloor \frac{f(1990)}{90} \right\rfloor \) can be either 21 or 22.

8. If \( \left\lfloor \frac{f(1990)}{90} \right\rfloor = 21 \):
   \[
   f(1990) = 14 + 90 \cdot 21 = 14 + 1890 = 1904
   \]

9. If \( \left\lfloor \frac{f(1990)}{90} \right\rfloor = 22 \):
   \[
   f(1990) = 14 + 90 \cdot 22 = 14 + 1980 = 1994
   \]

10. Both values \( 1904 \) and \( 1994 \) satisfy the condition \( 1900 < f(1990) < 2000 \).

The final answer is \( \boxed{1904, 1994} \).