122. The sum of several consecutive positive integers is 1000. Find these numbers.
122. We are given that $m+(m+1)+\ldots+(m+k)=1000$. By the formula for the sum of terms of an arithmetic progression, we have:

$$
\frac{2 m+k}{2} \cdot(k+1)=1000
$$

Or

$$
(2 m+k)(k+1)=2000
$$

Since

$$
(2 m+k)-(k+1)=2 m-1
$$

is odd, one of the last two factors is even, and the other is odd. Moreover, it is clear that $2 m+k>k+1$. From this, it follows that our problem has the following solutions:

$$
\begin{aligned}
2 m+k=2000, k+1=1, \quad m=1000, & k=0 \\
2 m+k=400, k+1=5, \quad m=108, & k=4 ; \\
2 m+k=80, k+1=25, m=28, & k=24 \\
2 m+k=125, k+1=16, m=55, & k=15
\end{aligned}
$$