Writing the expression in a common denominator gives $\frac{4a^2+2023b^2}{4ab}$ Let $\frac{4a^2+2023b^2}{4ab}=N$ where $N \in \mathbb{N}$. Notice that $b\mid4a^2$, and since we are given that the $\gcd(a,b)=1$, $b\mid4$. Thus, the possible values for $b$ are $1, 2,$ or $4$. By the same logic, $a\mid2023$ and therefore $a$ can be $1, 7, 17, 119, 289,$ or $2023$. Case 1: $b=1$ $\implies \frac{4a^2+2023}{4a}=N$ There are no solutions for this case since the numerator is odd while the denominator is even. Case 2: $b=2$ $\implies \frac{a^2+2023}{2a}=N$ For $a=1$, we have $N=1012$ For $a=7$, we have $N=148$ For $a=17$, we have $N=68$ For $a=119$, we have $N=68$ For $a=289$, we have $N=148$ For $a=2023$, we have $N=1012$ Case 3: $a=4$ $\implies \frac{a^2+8092}{4a}=N$ Similar to case 1, there are no solutions as $a^2$ is always odd and thus we have an odd numerator with an even denominator. Therefore, all the possible integer values are $\fbox{68, 148, 1012}$