7. Arrange the ten digits from 0 to 9 into a ten-digit number without repetition and with the first digit not being zero. The number of such ten-digit numbers that can be divided by 99 is $\qquad$.
7.285 120 .

Since the sum of the ten digits from 0 to 9 is $45$, and $9 \mid 45$, such a ten-digit number is always divisible by 9.

Now, divide the ten digits from 0 to 9 into two groups, each containing 5 digits, with their sums being $x$ and $y$ respectively, where $x \geqslant y \geqslant 10$.

If $11 \mid (x-y)$, then the ten-digit number is divisible by 11, with $x+y=45$, and $x-y=11k\left(k \in \mathbf{Z}_{+}\right)$.
Solving this, we get $x=28, y=17$.
Notice that, among the ten digits, there are 11 groups of five digits whose sum is 17:
$$
\begin{array}{l}
\{0,1,2,5,9\},\{0,1,2,6,8\}, \\
\{0,1,3,4,9\},\{0,1,3,5,8\}, \\
\{0,1,3,6,7\},\{0,1,4,5,7\}, \\
\{0,2,3,4,8\},\{0,2,3,5,7\}, \\
\{0,2,4,5,6\},\{1,2,3,4,7\}, \\
\{1,2,3,5,6\} .
\end{array}
$$

Therefore, the number of such ten-digit numbers is
$$
11 \times(5!\times 5!+4 \times 4!\times 5!)=285120 .
$$