25. Find the largest positive integer $m$, such that for $k \in \mathbf{N}^{*}$, if $1<k<m$ and $(k, m)=1$, then $k$ is a power of some prime.

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25. The required maximum positive integer $m=60$.

On the one hand, it can be directly verified that $m=60$ meets the requirement (it is only necessary to note that the product of the smallest two primes coprime with 60, 7 and 11, is greater than 60).

On the other hand, let $p_{1}, p_{2}, \cdots$ represent the sequence of primes in ascending order. Suppose $m$ is a positive integer that meets the requirement. We first prove: if $p_{n} p_{n+1} \leqslant n$, then
$$p_{1} p_{2} \cdots p_{n} \leqslant m$$

In fact, if there are two different numbers $p_{u}, p_{v}$ in $p_{1}, p_{2}, \cdots, p_{n+1}$ that are not divisors of $m$, then $\left(p_{u} p_{v}, m\right)=1, p_{u} p_{v} \leqslant p_{n} p_{n+1} \leqslant m$, which contradicts the property of $m$. Therefore, at most one of $p_{1}, p_{2}, \cdots, p_{n+1}$ is not a divisor of $m$, so (5) holds.

Now, if there exists $m(>60)$ that meets the requirement, then when $m \geqslant 77=p_{4} p_{5}$, let $n$ be the largest positive integer such that $p_{1} p_{2} \cdots p_{n} \leqslant m$. Then, by (5), we know $n \geqslant 4$. Combining the conclusion from the previous problem, we have $p_{n+1} p_{n+2}5 \times 7$, so $2,3,5,7$ have at most one that is not a divisor of $m$ (see the proof of (5)), thus $m$ is a multiple of one of the numbers 105, 70, 42, 30, hence $m=70$. But in this case, $(33, m)=1$, which contradicts the property of $m$.