$2 \cdot 60$ number $1978^{n}$ and $1978^{m}$ have the same last three digits, try to find the positive integers $m$ and $n$, where $n>m \geqslant 1$, such that $m+n$ takes the minimum value.
[Solution] Since the last three digits of $1978^{n}$ and $1978^{m}$ are the same, $1978^{n}-$ $1978^{m}=1978^{m}\left(1978^{n-m}-1\right)$ is a multiple of $1000=2^{3} \times 5^{3}$.

And $1978^{n-m}-1$ is an odd number, it has no factor of 2, so $1978^{m}$ needs to be divisible by $2^{3}$. 1978 has only one factor of 2, therefore $m \geqslant 3$.

Since $1978^{m}$ has no factor of 5, $1978^{n-m}-1$ needs to be divisible by $5^{3}$, the last digit of $1978^{t}$ is $8,4,2,6,8,4,2,6, \cdots$ repeating, only when $n-m=4 k$, $1978^{n-m}-1$ can be divisible by $5^{3}$.
Since $1978^{4 k}=(2000-22)^{4 k}$, it is sufficient that $22^{4 k}-1$ is divisible by $5^{3}$. And $22^{4 k}=484^{2 k}=(500-16)^{2 k}$, also $16^{2 k}=(250+6)^{k}$, therefore it is sufficient that $6^{k}-1$ is divisible by $5^{3}$.
$$
6^{k}-1=(6-1)\left(6^{k-1}+6^{k-2}+\cdots+6+1\right) \text {. }
$$

Therefore, $6^{k-1}+6^{k-2}+\cdots+6+1$ needs to be divisible by 25. Each term divided by 5 leaves a remainder of 1, there are $k$ terms, so $k$ should be a multiple of 5, i.e., $k=5 p$.
$$
\begin{aligned}
6^{k}-1 & =6^{5 p}-1=7776^{p}-1 \\
& =(7776-1)\left(7776^{p-1}+7776^{p-2}+\cdots+7776+1\right) .
\end{aligned}
$$
7775 is a multiple of 25, 7776 leaves a remainder of 1 when divided by 5, to make $7776^{p-1}+7776^{p-2}+\cdots+$ $7776+1$ divisible by 5, the minimum value of $P$ is $5, n-m=4 k=20 p$, so the minimum value of $n-m$ is 100.
To make $n+m$ the smallest, take $m=3$, then $n=103$.