2. Arrange all positive integers that are coprime with 105 in ascending order, and find the 1000th term of this sequence.
The problem is to find the number $n$ such that after removing the multiples of $3, 5, 7$ from $1$ to $n$, there are 1000 numbers left.
Thus, 
\[ n - \left( \left[ \frac{n}{3} \right] + \left[ \frac{n}{5} \right] + \left[ \frac{n}{7} \right] \right) + \left( \left[ \frac{n}{15} \right] + \left[ \frac{n}{21} \right] + \left[ \frac{n}{35} \right] \right) - \left[ \frac{n}{105} \right] = 1000 \]
\[
\Rightarrow \left( \left[ \frac{n}{3} \right] + \left[ \frac{n}{5} \right] + \left[ \frac{n}{7} \right] \right) - \left( \left[ \frac{n}{15} \right] + \left[ \frac{n}{21} \right] + \left[ \frac{n}{35} \right] \right) + \left[ \frac{n}{105} \right] = n - 1000 \text{. }
\]

Notice that 
\[ k - 1 \frac{71 n}{105} - 3 - \frac{15 n}{105} + \frac{n}{105} - 1 = \frac{19 n}{35} - 4 \]
\[
\Rightarrow \frac{16 n}{35} > 996 \Rightarrow n \geqslant 2179 \text{. }
\]

Since 
\[ \left( \left[ \frac{2184}{3} \right] + \left[ \frac{2184}{5} \right] + \left[ \frac{2184}{7} \right] \right) - \left( \left[ \frac{2184}{15} \right] + \left[ \frac{2184}{21} \right] + \left[ \frac{2184}{35} \right] \right) + \left[ \frac{2184}{105} \right] = 1185, \]
and 
\[ 2184 - 1185 = 999, \]
the 1000th number is 2186.

Thus, the 1000th term of this sequence is 2186.