N20 (20-1, Cuba) The last three digits of $1978^{n}$ and $1978^{m}$ are equal. Try to find positive integers $m$ and $n$, such that $m+n$ takes the minimum value (here $n>m \geqslant 1$).
Given the problem:
$$
\begin{aligned}
1978^{n}-1978^{m} & =1978^{m}\left(1978^{n-m}-1\right) \\
& \equiv 0 \quad(\bmod 1000) .
\end{aligned}
$$

Since \(2 \nmid\left(1978^{n-m}-1\right)\) and \(5 \nmid 1978^{m}\), we have
$$
\begin{array}{c}
1978^{m}=2^{m} \times 989^{m} \equiv 0 \quad(\bmod 8), \\
1978^{n-m} \equiv 1 \quad(\bmod 125) .
\end{array}
$$

From (1), we know \(m \geqslant 3\). From (2) and \(1978 \equiv 3(\bmod 5)\), we get \(3^{n-m} \equiv 1(\bmod 5)\). Since \(3^{n} \mod 5\) is: \(3,4,2,1,3,4,2,1, \cdots\), it follows that \(4 \mid n-m\). Let \(n-m = 4k\), then
$$
\begin{aligned}
1978^{n-m}-1 & \equiv(-22)^{4 k}-1 \\
& =(484)^{2 k}-1 \\
& =(500-16)^{2 k} \\
& \equiv 256^{k}-1 \\
& \equiv 6^{k}-1 \quad(\bmod 125) .
\end{aligned}
$$

And
$$
\begin{aligned}
6^{k}-1 & =(5+1)^{k}-1 \\
& \equiv 5^{2} \cdot \frac{k(k-1)}{2}+5 k \quad(\bmod 125),
\end{aligned}
$$

Thus,
$$
5^{2} \left\lvert\, 5 \cdot \frac{k(k-1)}{2}+k\right.,
$$

which means
$$
5^{2} \left\lvert\, \frac{k(5 k-3)}{2}\right. \text {. }
$$

Since \((5,5 k-3)=1\), it follows that \(5^{2} \mid k\), hence
$$
100 \mid n-m \text {. }
$$

Therefore, \(n-m \geqslant 100\), and \(m+n=n-m+2 m \geqslant 100+6=106\).
When \(m=3\) and \(n=103\), the conditions are satisfied, so the minimum value of \(m+n\) is 106.