Question 25. The sum of several consecutive natural numbers equals 1000, find these natural numbers.
Let the natural numbers be
$$
m+1, m+2, m+3, \cdots, m+n
$$
( $m, n$ are natural numbers). Then
$$
\frac{(m+1)+(m+n)}{2} \cdot n = 1000,
$$

which simplifies to $2 m n + n^{2} + n = 2000$.
Thus, $m = \frac{1000}{n} - \frac{n}{2} - \frac{1}{2}$.
Clearly, $m > 0$. Therefore, $\frac{1000}{n} > \frac{n}{2}$,

which implies $n^{2} < 2000$. Hence, $n < 45$.
From equation (1), $n$ can only take values of the form $2^{2} \cdot 5^{b}$ (where $a, b$ are integers). Solving (1) yields
$$
\left\{\begin{array}{l}
n = 5, \\
m = 197;
\end{array} \quad \left\{\begin{array}{l}
n = 16, \\
m = 54;
\end{array} \quad \left\{\begin{array}{l}
n = 25, \\
m = 27.
\end{array}\right.\right.\right.
$$

Thus, the required consecutive natural numbers are
$$
\begin{array}{l}
\text{198, 199, 200, 201, 202; } \\
\text{or} \\
55, 56, 57, \cdots, 70; \\
\text{or} \\
28, 29, 30, \cdots, 52.
\end{array}
$$