Solution We have $$n = 1 \times 3 \times 5 \times \cdots \times 2019 = \prod_{k = 0}^{1009} (2k + 1).$$ We wish to determine the last 3 digits, that is, $n \bmod{1000}$. By the Chinese remainder theorem, it suffices to consider this modulo 8 and modulo 125, as $\gcd(8, 125) = 1$, and $8 \times 125 = 1000$. Modulo 125, we have 125 being odd and between 1 and 2019, thus $125 \mid n$, and $n \equiv 0 \pmod{125}.$ Modulo 8, we write $$n = \prod_{k = 0}^{1009} (2k + 1) = 2017 \times 2019 \times \prod_{k = 0}^{251} (8k + 1)(8k + 3)(8k + 5)(8k + 7,$$ and thus $$\begin{align*} n &\equiv 1\times 3\times\prod_{k = 0}^{251}(1)(3)(-3)(-1) \\ &\equiv 3\times \prod_{k = 0}^{251} 9 \equiv 3\times \prod_{k = 0}^{251} 1 \\ &\equiv 3 \pmod{8}. \end{align*}$$ Thus if $x$ is the last 3 digits, then $x$ is the least solution of $$\left.\begin{array}{l} x \equiv 0 \; \bmod{125} \\ x \equiv 3 \; \bmod{8} \end{array}\right\}.$$ Now $x \equiv 0 \pmod{125} \implies x = 125 a,$ $a \in \mathbb{N}$. Then $125 a \equiv 3 \pmod{8} \implies 5a \equiv 3 \pmod{8}$. Solving, we have $a = 7$, and thus $x = 125 \cdot 7 = 875$, so the last 3 digits of $n$ are 875.