At least $86$ rows are needed.
Suppose there are $86$ (type A) countries with $23$ representatives each, $10$ countries with $2$ representatives each, and $4$ countries with $1$ representative each. Then there are $86 + 10 + 4 = 100$ countries and $86 \times 23 + 10 \times 2 + 4 \times 1 = 2002$ representatives. No two type A countries may sit in the same row since $23 \times 2 > 45$. Thus, at least $86$ rows are needed.

We now show that it is always sufficient to use $86$ rows. We arrange the countries in decreasing order of the number of representatives. We first put all representatives from country $1$ in the first row. Next, if all representatives from country $2$ can be seated in the same row as country $1$, we put them in that row, or otherwise they are put in the next row. Similarly, if all representatives from country $k+1$ can be seated in the same row as country $k$, we put them in that row, or otherwise they are put in the next row. We do the same thing until all representatives are seated or all $86$ rows are used up.

It suffices to consider the case when all $86$ rows are used up but some representatives still do not have a seat. We claim that each row consists of at least $23$ representatives. If not, there are at least $23$ empty seats in some row $j$, which means the next country to be put has at least $24$ representatives. But then there should be at least $24$ representatives in row $j$ as well according to the order of the countries. This is a contradiction. It follows that the number of representatives remaining is at most
$$2002 - 23 \times 86 = 24.$$
Among these representatives, suppose $c$ of them come from the same country, where $c$ is maximized. First of all, we have $c \le 22$ since otherwise at least $99$ countries have at least $23$ representatives (except possibly the last country), which yields the contradiction that the total number of representatives is at least
$$23 \times 99 = 2277 > 2002.$$
Now, we partition the remaining representatives into at most two groups by putting those $c$ representatives in one group. We claim that all groups have at most $22$ representatives. If not, since $c \le 22$, it must be the case that $c = 1$ and there are $23$ more representatives. By the choice of $c$, all these $24$ representatives come from different countries. But this is impossible since there are at most $100 - 86 = 14$ countries left.

Therefore, it remains to show that there are at least $2$ rows with $22$ empty seats. Indeed, if this does not hold, then the total number of representatives is at least
$$24 \times 85 + 23 = 2063 > 2002.$$
This is a contradiction, and so we are done.