18. Let $m, n (m>n)$ be positive integers, and $70^{2} \mid 2023^{m}-2023^{n}$, then the minimum value of $m+n$ is $\qquad$
$2023=7 \times 17^{2}, 70^{2} \mid 2023^{m}-2023^{n}=2023^{n}\left(2023^{m-n}-1\right)$. Since $(2023,100)=1$, then $\left\{\begin{array}{l}7^{2} \mid 2023^{n}, \\ 100 \mid 2023^{m-n}-1\end{array} \Rightarrow\left\{\begin{array}{l}n \geqslant 2, \\ 2023^{m-n} \equiv 1(\bmod 100) \text {. }\end{array}\right.\right.$
Also, $\varphi(100)=40$, by Euler's theorem, $2023^{40} \equiv 1(\bmod 100)$, thus $2023^{k} \equiv 1(\bmod 100) \Rightarrow k \mid 40$. Calculation yields $2023^{20} \equiv 23^{20} \equiv 29^{10} \equiv 41^{5} \equiv 41 \cdot 81^{2}=41 \cdot 61 \equiv 1(\bmod 100)$, $2023^{10} \equiv 23^{10} \equiv 29^{5} \equiv 29 \cdot 41^{2} \equiv 29 \cdot 81 \equiv 49(\bmod 100)$, hence $k=20 t\left(k \in \mathbf{N}^{*}\right)$.

In summary, $m+n=m-n+2 n \geqslant 20+2 \times 2=24$.