Initially, the level markers for internet users are stars, moons, and suns. The rule is: if the cumulative online time from 0:00 to 24:00 in a day is full 2 hours, it counts as 1 day of internet use. Starting from the first day of internet use, after 5 days, the level marker is 1 star; after another 7 days, the level marker is 2 stars; ... (to get the $n$-th star, it takes 2 more days than to get the $(n-1)$-th star). Every 4 stars are equivalent to 1 moon, and every 4 moons are equivalent to 1 sun (i.e., 16 stars equal 1 sun). If starting from July 1, 2005, and using the internet every day without interruption, what will the level marker be on January 1, 2011? How many more days will it take to enter the next level marker?
Solving, given that July 1, 2005, is the 1st day of internet usage, we know that January 1, 2011, is the
$$
\begin{array}{l}
185+365 \times 5+1 \text { (2008 is a leap year, with } 366 \text { days) } \\
=2011 \text { days. }
\end{array}
$$

Let the level marker on January 1, 2011, be $n$ stars. We know that $n$ is the maximum value satisfying
$$
5+7+\cdots+(2 n+3)<2011
$$

By
$$
\begin{array}{l}
\text { } \frac{n(5+2 n+3)}{2}<2011, \text { i.e., } \\
n^{2}+4 n-2011<0,
\end{array}
$$

we solve to get $n \leqslant 42$.
Given $42=16+16+4+4+1+1$, we know that the level marker on January 1, 2011, is 2 suns, 2 moons, and 2 stars.
$$
\text { By } \frac{43 \times(5+2 \times 43+3)}{2}-2011=10 \text {, we }
$$

know that it will take another 10 days to enter the next level marker.