We have $20n+2\mid 2023n+210 \rightarrow 20n+2\mid 3n+8$ $\rightarrow 20n+2 \le 3n+8 \rightarrow n=0$

This is Easy Since, \[ \frac{2023n + 210}{20n+2} = \frac{101(20n+2) + 3n + 8}{20n+2} \implies \frac{3n + 8}{20n+2} \in \mathbb{Z} \]This tells us, \[ 20n + 2 \leq 3n + 8 \implies n \leq \frac{6}{17} \] Since, We are given $n \geq 0$ Only Solution is $\boxed{n = 0}$