Let N be the set of all natural numbers and$f:N\to N$ be such that $1990<f\left(1990\right)<2100$ and satisfies the equation $x-f\left(x\right)=19\left[\frac{x}{19}\right]-90\left[\frac{f\left(x\right)}{90}\right]$ where $\left[y\right]$ denotes the greatest integer less than or equal to $y$. then the number of possible values of $f\left(1990\right)$ is 

The given function is $x-f\left(x\right)=19\left[\frac{x}{19}\right]-90\left[\frac{f\left(x\right)}{90}\right]$ and the condition is $1990<f\left(1990\right)<2100$

we know that $\left[\frac{1990 to 2100}{90}\right]=22 or 23$

Subsitute $x=1990$in the function

Case (I)

$\begin{array}{c}1990-f\left(1990\right)=19\left[\frac{1990}{19}\right]-90\left[\frac{f\left(1990\right)}{90}\right]\\ 1990-f\left(1990\right)=19\left(104\right)-90\left(22\right)\end{array}$

this will give one value 

Case (II)

$\begin{array}{c}1990-f\left(1990\right)=19\left[\frac{1990}{19}\right]-90\left[\frac{f\left(1990\right)}{90}\right]\\ 1990-f\left(1990\right)=19\left(104\right)-90\left(23\right)\end{array}$

This gives second value 

Therefore, the number of values for $f\left(1990\right)$ is $\boxed{2}$