First, let's establish a lower bound on $n$.
A polyomino with an area of $k$ squares can cover at most $k+1$ rows and columns. This is because a polyomino is a connected region, and if you add a square to the polyomino to enlarge it, it may expand but only in at most one direction, encroaching on just one additional row or column. A monomino covers $2$ ($1$ row and $1$ column) so a $k$-omino covers at most $k+1$ rows/columns.
For each $k\in\{1,2,...,9\}$, we need the $k$-ominos to cover the $2n$ rows/columns of the $n\times n$ board. Each one covers at most $k+1$, so we need at least $\lceil \frac{2n}{k+1} \rceil$ of them. Adding together their total area, we find that it exceeds $n^2$, the area of the board, for all $n<15$.
So we have a lower bound of $n=15$. Here is a table of the amounts in this case, and it turns out that the pieces would have to completely fill the board:
k #pcs Area
1 15 15
2 10 20
3 8 24
4 6 24
5 5 25
6 5 30
7 4 28
8 4 32
9 3 27
---
Total: 225
This initial lower bound can be improved upon.
With a large polyomino you have some leeway in how many rows/columns it will cover. For example, the 9-omino can cover just 6 rows/columns if it is a 3x3 square, or 10 rows/columns when it is straight. Smaller polyominoes are less flexible.
The domino always covers exactly 3 rows/columns. For this puzzle that means that $2n$ is a multiple of $3$, and hence that $n$ is a multiple of $3$.
Similarly, the tromino always covers exactly 4 rows/columns, whichever shape it is. This means that $2n$ must be a multiple of $4$, and hence that $n$ is even.
Given these last two requirements, the smallest board size that might work is $n=18$.
Here is a solution which is highly likely to be non-optimal. I hope others can find a solution that attains the lower bound established above.
Here $n=24$. It is easy to use a $k$-omino to cover all the rows/columns of a $3\times3$ or $2\times2$ square. I combine some of these so as to make four kinds of $6\times6$ square such that each row/column is covered by the $k$-ominos being used. I then combine these four types of $6\times6$ square so that each row/column has one of each type.
Edit:
Retudin has found a solution that does attain the lower bound, and which is therefore optimal.