Let
$$f_{m, n}(i)=\left[\frac{i}{m}\right]+\left[\frac{i}{n}\right], S(m, n)=\sum_{i=0}^{m n-1}(-1)^{f_{m, n} n^{(i)}},$$

The problem is equivalent to finding the necessary and sufficient condition for $S(m, n)=0$.
Case 1: Both $m$ and $n$ are odd. In this case, $S(m, n)$ is the sum of an odd number $(m n)$ of odd numbers, so $S(m, n) \neq 0$.

Case 2: One of $m$ and $n$ is odd, and the other is even. In this case, for $0 \leqslant i \leqslant m n-1$, we have
$$\begin{aligned} f_{m, n}(m n-i-1) & =\left[\frac{m n-i-1}{m}\right]+\left[\frac{m n-i-1}{n}\right] \\ & =m+n+\left[-\frac{i+1}{m}\right]+\left[-\frac{i+1}{n}\right] \end{aligned}$$

We need the following result: For $m \in \mathbf{N}^{*}, i \in \mathbf{N}$,
$$\left[\frac{i}{m}\right]+\left[-\frac{i+1}{m}\right]=-1$$

In fact, let $\frac{i}{m}=k+\alpha, k \in \mathbf{N}, 0 \leqslant \alpha<1$, then $\left[\frac{i}{m}\right]=k$, and $0 \leqslant \alpha \leqslant \frac{m-1}{m}$. At this time,
$$-\frac{i+1}{m}=-k-\alpha-\frac{1}{m} \geqslant-k-\frac{m-1}{m}-\frac{1}{m}=-k-1,$$

and
$$-\frac{i+1}{m}=-k-\alpha-\frac{1}{m} \leqslant-k-\frac{1}{m}<-k$$

Therefore,
$$\left[-\frac{i+1}{m}\right]=-(k+1)$$

This shows that (8) holds.
From (8) and (7), we know
$$\begin{aligned} f_{m, n}(m n-i-1) & =m+n-1-\left[\frac{i}{m}\right]-1-\left[\frac{i}{n}\right] \\ & =m+n-2-f_{m, n}(i) \\ & \equiv f_{m, n}(i)+1(\bmod 2) \end{aligned}$$

Thus, by pairing the first and last terms in the sum $S(m, n)$ (noting that the number of terms $m n$ is even), we can see that $S(m, n)=0$.

Case 3: Both $m$ and $n$ are even. Let $m=2 k, n=2 l$. Since for any $p \in \mathbf{N}^{*}, j \in \mathbf{N}$, we have $\left[\frac{2 j}{2 p}\right]=\left[\frac{2 j+1}{2 p}\right]$, therefore, $S(m, n)$ is twice the sum of all even $i (0, 2, \cdots, m n-2)$, i.e.,
$$\begin{aligned} S(m, n) & =2 \sum_{i=0}^{2 H-1}(-1)^{f_{m, n^{(i}}}=2 \sum_{i=0}^{2 H-1}(-1)^{f_{k, l^{(i)}}} \\ & =2\left(S(k, l)+\sum_{i=k l}^{2 H-1}(-1)^{f_{k, l^{(i)}}}\right) \\ & =2\left(S(k, l)+\sum_{i=0}^{k-1}(-1)^{k+l} \cdot(-1)^{f_{k, l^{(i}}(i)}\right) \\ & =2 S(k, l)\left(1+(-1)^{k+l}\right) \end{aligned}$$

By replacing $k, l$ with $m, n$ and repeating the above discussion until $k, l$ are not both even, we find that the necessary and sufficient condition for $S(m, n)=0$ is: the exponents of 2 in the prime factorizations of $m$ and $n$ are different.