Solution:
Yes. Place a grid of $900$ points in $30$ equally spaced rows and columns, so that each point is a distance $15/31$ from its nearest neighbours (or $15/31$ from the edge). This blocks all rectangles except those slimmer than $1/2$. Those slimmer than $1/2$ must have length at least $2$, so we can block them with a smaller set of rows and columns containing more finely spaced points.

Label the rows $1$-$30$. In each of the $7$ rows $3$, $7$, $11$, $15$, $19$, $23$, $27$ place an additional $31$ points, so that each of these rows has $61$ equally spaced points at a spacing of $15/62$. Similarly for the columns. So in total we are placing an additional $2 \times 7 \times 31 = 434$ points. Any rectangle of length $>2$ must encounter one of these rows (or columns) and hence must have width less than $1/4$. This blocks any rectangle except those with width $< 1 / 4$.

In each of the $3$ rows $7$, $15$, $23$ place an additional $62$ points, so that each of these rows has $123$ equally spaced points at a spacing of $15/124$. Similarly for the columns. So in total we are placing an additional $2 \times 3 \times 62 = 372$ points. Any rectangle of length $>4$ must encounter one of these rows (or columns) and hence must have width less than $1/8$. This blocks any rectangle except those with width $< 1 / 8$ and hence length $>8$.

In row $15$ place an additional $124$ points, so that it has a total of $247$ equally spaced points at a spacing of $15/247$. Similarly for column $15$. This requires an additional $248$ points. Any rectangle which can fit through these gaps has area at most $15 \times 15 / 247 < 1$. So we have blocked all rectangles with area $1$ or more and used $900 + 434 + 372 + 248 = 1954$ points.

Ilan Mayer, who seems to solve these problems effortlessly, came up with a neater arrangement of points. He used narrowly spaced points along widely spaced diagonals: $(k/15, k/15)$ for $k = 1,2,\ldots ,224$; $((28n+k)/15, k/15)$ for $n = 1,2,\ldots ,7$, $k = 1,2,\ldots ,224 - 28n$; $(k/15, (28n + k)/15)$ for $n = 1,2,\dots,7$, $k = 1,2,\dots,224 - 28n$. The diagonals are spaced $28/15$ apart, so the biggest rectangle that can be fitted between two diagonals has sides $15/15$ less $\varepsilon$ and $15/15$ less $\varepsilon$. For example, take the vertices as $(14 / 15 + \varepsilon,\varepsilon)$, $(29 / 15 - \varepsilon,\varepsilon)$, $(14 / 15 + \varepsilon, 15 / 15 - \varepsilon)$, $(29 / 15 - \varepsilon,15 / 15 - \varepsilon)$. If one allows a rectangle to touch points (in other words if one took the rectangles to exclude their boundaries) then this does not work - many $15 \times 1 / 15$ rectangles will fit. But one can add an additional point on each of the $15$ lines, keeping the points on each line evenly spaced. That blocks rectangles without boundary and still has only $1821$ points.