is it possible to divide $9\times 11$ rectangle into one $1\times 3$ rectangle and N tetrominoes?

Here's what turns out to be the only solution, with the $1 \times 3$ tromino in the center:

   i\j 1  2  3  4  5  6  7  8  9 10 11 
    1  2  2  2  3  4  4  4  5  6  6  6 
    2 18  2  3  3  3  4  5  5  5  6 23 
    3 18 18  7  7  7  8  9  9  9 23 23 
    4 18 19 21  7  8  8  8  9 22 24 23 
    5 19 19 21 21  1  1  1 22 22 24 24 
    6 20 19 21 10 11 11 11 12 22 24 25 
    7 20 20 10 10 10 11 12 12 12 25 25 
    8 20 13 14 14 14 15 16 16 16 17 25 
    9 13 13 13 14 15 15 15 16 17 17 17 

I used the following integer linear programming formulation. Let $P$ be the set of polyominoes, and let $T \subset P$ be the set of $1 \times 3$ trominoes. Note that $|T| = 9\cdot 9+ 7\cdot 11=158$ and $|P|=|T|+ 2(8\cdot 9+7\cdot 10)= 442$. For $i\in\{1,\dots,9\}, j\in\{1,\dots,11\}$, let $P_{i,j}\subset P$ be the subset of polyominoes that contain square $(i,j)$. Let binary decision variable $x_p$ indicate whether polyomino $p\in P$ is used. The constraints are: \begin{align} \sum_{p \in P_{i,j}} x_p &= 1 &&\text{for $i\in\{1,\dots,9\}, j\in\{1,\dots,11\}$} \\ \sum_{t\in T} x_t &= 1 \\ x_p &\in \{0,1\} &&\text{for $p \in P$} \end{align} The first constraint enforces that exactly one polyomino contains square $(i,j)$. The second constraint forces exactly one tromino to be used.

Here is a picture which proves it is possible.