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.