25. For $n \in \mathbf{N}^{*}$, let $f(n)$ denote the smallest positive integer such that: $n \mid \sum_{k=1}^{f(n)} k$. Find all $n \in \mathbf{N}^{*}$ such that $f(n)=2 n-1$.
25. First, we prove: If \( n=2^{m}, m \in \mathbf{N} \), then \( f(n)=2 n-1 \).

In fact, on one hand,
\[
\sum_{k=1}^{2 n-1} k=(2 n-1) n=\left(2^{m+1}-1\right) 2^{m}
\]
is divisible by \( n \). On the other hand, if \( l \leqslant 2 n-2 \), then
\[
\sum_{k=1}^{l} k=\frac{1}{2} l(l+1)
\]
Since one of \( l \) and \( l+1 \) is odd, and
\[
l+1 \leqslant 2 n-1=2^{m+1}-1,
\]
the above sum cannot be divisible by \( 2^{m} \) (because \( 2^{m+1} \times l(l+1) \)).
Next, we prove: When \( n \) is not a power of 2, \( f(n) < 2 n-1 \). Let \( n=2^{m} p \), where \( p \) is an odd number. We show that there exists \( l < 2 n-1 \) such that \( 2^{m+1} \mid l \) and \( p \mid (l+1) \) (in this case, of course, \( 2^{m} p \left\lvert\, \frac{l(l+1)}{2}\right. \), thus \( f(n) < 2 n-1 \)).

Since \( \left(2^{m+1}, p\right)=1 \), by the Chinese Remainder Theorem,
\[
l \equiv 0\left(\bmod 2^{m+1}\right), \quad l \equiv p-1(\bmod p),
\]
has a solution \( l \equiv x_{0}\left(\bmod 2^{m+1} p\right) \). Therefore, there exists \( l_{0}, 0 < l_{0} \leqslant 2^{m+1} p \) satisfying the above system of congruences. Note that \( 2 n-1 \neq 0\left(\bmod 2^{m+1}\right) \), and \( 2 n+1 \not \equiv 0(\bmod p) \), so \( 2 n-1 \) and \( 2 n \) are not solutions to this system of congruences, hence \( 0 < l_{0} < 2 n-1 \), i.e., \( f(n) < 2 n-1 \).

In summary, \( f(n) = 2 n-1 \) if and only if \( n \) is a power of 2.