, 2024

Solution:
We simplify the expression as follows:
$$\begin{aligned} \frac{(2024 n)(2024 n+1) / 2}{(4 n)(4 n+1) / 2} & = \frac{506 \cdot (2024 n+1)}{4 n+1} \\ & = \frac{506 \cdot (506 \cdot (4 n+1) - 505)}{4 n+1} \\ & = 506^{2} - \frac{506 \cdot 505}{4 n+1} \\ & = 506^{2} - \frac{2 \cdot 5 \cdot 11 \cdot 23 \cdot 101}{4 n+1} . \end{aligned}$$

Thus, the expression is an integer if and only if $4 n+1$ divides $5 \cdot 11 \cdot 23 \cdot 101$. The smallest divisors of $5 \cdot 11 \cdot 23 \cdot 101$ are
$$1, 5, 11, 23, 55, 101, 253 .$$
Since $4 n+1 > 1$ and is $1$ modulo $4$, the three smallest values it can take are $5$, $101$, and $253$. Hence, the three smallest values of $n$ are $1$, $25$, and $63$, giving the answer of $1+25+63=89$.