10.5. Let $k$ be a positive integer. An infinite strictly increasing sequence of positive integers $\left\{a_{n}\right\}$ satisfies: for any positive integer $n$, we have $a_{n+1}-a_{n} \leqslant k$, and $a_{n}$ is a multiple of 1005 or 1006, but not a multiple of 97. Find the smallest possible value of $k$.
10.5. $k_{\text {min }}=2010$.

Given a positive integer $N$, such that
$$
a_{1}<1005 \times 1006 \times 97 \times N=D \text {. }
$$

Since $D$ is a multiple of 97, it is not any term in the sequence $a_{n}$, hence there exists a positive integer $n$, such that $a_{n}<D<a_{n+1}$.

The largest multiple of 1005 or 1006 less than $D$ is $D-1005$, and the smallest multiple of 1005 or 1006 greater than $D$ is $D+1005$, which implies
$$
\begin{array}{l}
a_{n} \leqslant D-1005, a_{n+1} \geqslant D+1005 . \\
\text { Hence } k \geqslant a_{n+1}-a_{n} \\
\geqslant(D+1005)-(D-1005)=2010 .
\end{array}
$$

Let $\left\{a_{n}\right\}$ be the increasing sequence of all multiples of 1005 that are not multiples of 97. Then
$$
\begin{array}{l}
a_{n+1}-a_{n} \leqslant 2010 . \\
\text { Hence } k_{\min }=2010 .
\end{array}
$$