Suppose that $f_{1}(x)=\frac{1}{2-x}$. For each positive integer $n \geq 2$, define $f_{n}(x)=f_{1}\left(f_{n-1}(x)\right)$ for all real numbers $x$ in the domain of $f_{1}\left(f_{n-1}(x)\right)$. The value of $f_{2019}(4)$ can be written as $\frac{a}{b}$ where $a$ and $b$ are positive integers with no common divisor larger than 1 . What is $(a, b) ?$
Since $f_{1}(x)=\frac{1}{2-x}$, then $f_{1}(4)=\frac{1}{2-4}=-\frac{1}{2}$.

Since $f_{2}(x)=f_{1}\left(f_{1}(x)\right)$, then $f_{2}(4)=f_{1}\left(-\frac{1}{2}\right)=\frac{1}{2-(-1 / 2)}=\frac{1}{5 / 2}=\frac{2}{5}$.

Since $f_{3}(x)=f_{1}\left(f_{2}(x)\right)$, then $f_{3}(4)=f_{1}\left(\frac{2}{5}\right)=\frac{1}{2-(2 / 5)}=\frac{1}{8 / 5}=\frac{5}{8}$.

Since we can re-write $f_{1}(4)=\frac{-1}{2}$, then the values $f_{1}(4), f_{2}(4), f_{3}(4)$ each satisfy the equality $f_{n}(4)=\frac{3 n-4}{3 n-1}$.

Suppose that $f_{k}(4)=\frac{3 k-4}{3 k-1}$.

Then

$$
\begin{aligned}
f_{k+1}(4) & =f_{1}\left(f_{k}(4)\right) \\
& =f_{1}\left(\frac{3 k-4}{3 k-1}\right) \\
& =\frac{1}{2-(3 k-4) /(3 k-1)} \\
& =\frac{3 k-1}{2(3 k-1)-(3 k-4)} \\
& =\frac{3 k-1}{3 k+2} \\
& =\frac{3(k+1)-4}{3(k+1)-1}
\end{aligned}
$$

This means that if one term in the sequence $f_{1}(4), f_{2}(4), f_{3}(4), \ldots$ is of this form, then the next term does, which means that all terms are of this form.

Therefore, $f_{2019}(4)=\frac{3(2019)-4}{3(2019)-1}=\frac{6053}{6056}$.

Since $6056-6053=3$, then any integer that is a divisor of both 6056 and 6053 is also a divisor of 3 .

This means that the only possible positive common divisors of 6053 and 6053 are 1 and 3.

Since neither 6053 nor 6056 is divisible by 3, then they have no common divisor greater than 1 . Thus, if $f_{2019}(4)=\frac{a}{b}$ where $a$ and $b$ are positive integers with no common divisor larger than 1 , then $(a, b)=(6053,6056)$.

ANSWER: $(6053,6056)$