We have \begin{align*} 2023 &= qn + \left(223- \frac{3}{2} \cdot n\right) \\ \iff 1800 &= \left(q - \frac{3}{2}\right)n \\ \iff 3600 &= (2q-3)n. \end{align*}Since $3600$ is a perfect square and $n$ is also a perfect square, therefore we must have $2q-3$ is a perfect square. Note that $3600=2^4\cdot 3^2\cdot 5^2$, therefore $2q-3\in\{2^2, 2^4, 2^2\cdot 3^2, 2^2\cdot 5^2, 2^4\cdot 3^2, 2^4\cdot 5^2, 2^2\cdot 3^2\cdot 5^2, 2^4\cdot 3^2\cdot 5^2, 3^2, 5^2, 3^2\cdot 5^2\}$. Note that $2q-3$ is an odd number. $223- \frac{3}{2} \cdot n$ is the remainder of the division, hence $0\le 223- \frac{3}{2} \cdot n\le n$. The left inequality is equivalent to $\frac{3}{2} \cdot n\le 223$, or $n\le 148$. The right inequality is equivalent to $223\le\frac{5}{2}n$, or $n\ge 89$. These inequality leads to $\left\lceil\frac{3600}{148}\right\rceil \le 2q-3 \le \left\lfloor\frac{3600}{89}\right\rfloor$, or $25\le 2q-3\le 40$. From these two observations, we have $2q-3=5^2$, from which we have $n=144$, $q=14$.