8.3. Let $p(n, k)$ denote the number of divisors of the positive integer $n$ that are not less than $k$. Try to find
$$
\begin{array}{l}
p(1001,1)+p(1002,2)+\cdots+ \\
p(2000,1000) .
\end{array}
$$
8.3. If we write down all the divisors of each positive integer $1000+k$ that are not less than $k$ $(k=1,2, \cdots, 1000)$, then the sum we are looking for is the sum of the number of all these divisors.

Below we prove that each positive integer $n$ from 1 to 2000 is written exactly once. If $n>1000$, then it is only written when it is a divisor of itself. If $n \leqslant 1000$, then for $k \leqslant n$, when $1000+k$ is a multiple of $n$, it is listed among the divisors of $1000+k$. Since there is exactly one number that is a multiple of $n$ between $1000+1$ and $1000+n$, $n$ is also written exactly once. Therefore, each positive integer $n$ from 1 to 2000 is written exactly once, and they total 2000.