18. 11 positive integers are arranged in a sequence, with their mean, median, and mode being $10$, $9$, and $8$ respectively. The maximum possible value of the largest number in this sequence is ( ).
(A) 24
(B) 30
(C) 31
(D) 33
(E) 35
18. E.

Since 9 is the median, let the 11 positive integers in ascending order be
$a, b, c, d, e, 9, f, g, h, i, j$.
Since 8 is the mode, 8 must appear at least twice.
We will consider different cases.
(1) When 8 appears twice, let the 11 numbers be
$a, b, c, 8, 8, 9, f, g, h, i, j$.
To make the maximum number $j$ as large as possible, the other numbers should be as small as possible, so let the 11 numbers be
$$
\begin{array}{l}
1, 2, 3, 8, 8, 9, 10, 11, 12, 13, j. \\
\text { Hence } j=11 \times 10-(1+2+3+8+8+9+ \\
10+11+12+13) \\
=33 .
\end{array}
$$
(2) When 8 appears three times, let the 11 numbers be
$$
a, b, 8, 8, 8, 9, f, g, h, i, j.
$$

To make the maximum number $j$ as large as possible, the other numbers should be as small as possible, so let the 11 numbers be
$$
\begin{array}{l}
1, 1, 8, 8, 8, 9, 9, 10, 10, 11, j. \\
\text { Hence } j=11 \times 10-(1+1+8+8+8+9+9+ \\
10+10+11) \\
=35 .
\end{array}
$$

It is easy to verify that when 8 appears more than three times, $j < 35$.
In summary, the maximum possible value of the largest number in this sequence is 35.