The last three digits of $1978^{n}$ and $1978^{m}$ are equal. Try to find positive integers $n, m$, such that $m+n$ takes the minimum value. Here $n>m \geqslant 1$.


Given that $1978^{n}-1978^{m}=1978^{m}\left(1978^{n-m}-1\right) \equiv 0(\bmod 1000)$, therefore
$$
1978^{m}=2^{m} \times 989^{m} \equiv 0(\bmod 8), 1978^{n-m} \equiv 1(\bmod 125).
$$

Hence $m \geqslant 3$. Also,
$$
\begin{aligned}
(1978)^{n-m} & =(1975+3)^{n-m} \\
& \equiv 3^{n-m}+(n-m) 3^{n-m-1} \cdot 1975(\bmod 125),
\end{aligned}
$$

thus $3^{n-m} \equiv 1(\bmod 5)$, so $4 \mid n-m$.
Let $n-m=4 k(k \in \mathbf{N})$, then $3^{n-m}=81^{k} \equiv 1+80 k+\frac{k(k-1)}{2} \cdot 80^{2}(\bmod 125)$, substituting into (1) and (2) we get $1=1978^{4 k}=1+80 k+\frac{k(k-1)}{2} \cdot 80^{2}+4 k \cdot 1975(\bmod 125)$, thus $k(20 k+3) \equiv 0(\bmod 25)$, hence $25 \mid k, n-m$ is at least $4 \times 25=100$, therefore the minimum value of $m+n$ is $n-m+2 m=106$, here $m=3, n=103$.