We will show that $N = k^2$ is the desired least number for any $kn \times kn$ board tiled with $1 \times n$ tiles, $n \ge 2$, thus the answer to our problem is $N = 100^2 = 10000$. From now on consider a $kn \times kn$ board tiled with $1 \times n$ tiles ($n \ge 2$), with $k > 1$ fixed.

For a lower bound of $N$, divide the board into $k^2$ squares of size $n \times n$ and notice that if one of them, say $P$, does not contain some one of the $N$ chosen cells, then there exist two tilings which differ on $P$ but agree on the rest of the board, and the given tiling would never be uncovered. Thus we must have $k^2 \le N$.

For an upper bound of $N$, consider the $k^2$ in number lower-left cells in each one of the squares of size $n \times n$ considered above, and learn the position of the tiles that cover these cells. We shall show by induction on $n$ that this procedure allows you to figure out the complete tiling. Thus $N \le k^2$. This, together with $k^2 \le N$ prove above would imply $N = k^2$, and we'll have finished.

For the induction argue as follows:
For $n = 2$, suppose that there are two tilings $A$ and $B$ that agree on all lower-left cells of the $n \times n = 2 \times 2$ squares considered above, but differ on some other cell $a_1$ of the given board. We'll arrive at a contradiction, which will mean any tiling can be uncovered by choosing to learn the positions of these lower-left cells, as wanted. Indeed:
Let $d_1$ be the $1 \times n = 1 \times 2$ domino which covers $a_1$ in $A$. Colour in red the upper-right cell in each $2 \times 2$ square; without loss of generality, $a_1$ is a red cell (otherwise, replace $a_1$ by the other cell covered by $d_1$). Let $d_2$ be the $1 \times n = 1 \times 2$ domino which covers $a_1$ in $B$, $b_1$ be the second cell covered by $d_2$, $d_3$ be the domino which covers $b_1$ in $A$, $a_2$ be the second cell covered by $d_3$, et cetera. Then $a_i$ is red for all $i$.
Let $s$ be the least positive integer such that there exists a $t < s$ such that $a_s = a_t$; clearly, we must have $t = 1$. Consider the polyomino $P$ enclosed by the sequence of cells $a_1, b_1, a_2, b_2, \dots, b_{s-1}$, $a_s = a_1$.
Let $O_i$ be the centre of $a_i$. It is straightforward to verify that $S(P) = S(O_1O_2\dots O_{s-1}) - s + 2$ (where $S(\cdot)$ denotes area). Since $O_1O_2\dots O_{s-1}$ is a polyomino in the grid formed by the centres of all red cells, its area $S(O_1O_2\dots O_{s-1})$ is a multiple of four and the number $s-1$ of its sides is even. It follows from this that $S(P)$ is odd whereas it must be possible to tile $P$ with dominoes; the desired contradiction.

For the inductive step, let $n \ge 3$. Number all columns as $1$ through $kn$ from left to right and all rows as $1$ through $kn$ from bottom to top, and delete all rows and columns whose number is congruent to $2$ modulo $n$. This operation produces a $k(n-1) \times k(n-1)$ chessboard tiled with $1 \times (n-1)$ tiles. By the induction hypothesis, we can reconstruct this tiling completely by choosing to uncover the lower-left cells of the $(n-1) \times (n-1)$ squares in which we divide this $k(n-1) \times k(n-1)$ chessboard. These cells are exactly the lower-left cells of the $n \times n$ squares in which we divide the original $kn \times kn$ chessboard. Restore all deleted rows and columns, and you have the positions of all tiles in the original tiling that were not contained within a deleted row or column. Repeat the operation for all rows and columns whose numbers are congruent to $3$ modulo $n$, and you have the positions of those tiles as well, using the same lower-left cells of the $n \times n$ squares in which we divide the original $kn \times kn$ chessboard. So choosing to learn the positions of the tiles covering these cells you uncover the whole tiling of the original $kn \times kn$ chessboard, as wanted to prove. $\square$