10. For $i=2,3, \cdots, k$, the remainder when the positive integer $n$ is divided by $i$ is $i-1$. If the smallest value of $n$, $n_{0}$, satisfies $2000<n_{0}<3000$, then the smallest value of the positive integer $k$ is
10.9.

Since $n+1$ is a multiple of $2,3, \cdots, k$, the smallest value of $n$, $n_{0}$, satisfies
$$
n_{0}+1=[2,3, \cdots, k],
$$

where $[2,3, \cdots, k]$ represents the least common multiple of $2,3, \cdots, k$.
$$
\begin{array}{l}
\text { Since }[2,3, \cdots, 8]=840, \\
{[2,3, \cdots, 9]=2520,} \\
{[2,3, \cdots, 10]=2520,} \\
{[2,3, \cdots, 11]=27720,}
\end{array}
$$

Therefore, the smallest positive integer $k$ that satisfies $2000<n_{0}<3000$ is 9.