Solution:

If a circle with unit diameter intersects a unit square, then its center must lie inside an area $3 + \pi /4$, namely an oval centered on the square and comprising: the original square, area $1$; four $1 \times 1/2$ rectangles on the sides, total area $2$; and four quarter circles at the corners, total area $\pi /4$. So if it does not intersect any of the $120$ unit squares, then it must avoid ovals with a total area of $120 \times (3 + \pi /4) = 454.2$. Of course, for many arrangements of the squares, these ovals might overlap substantially, but the worst case would be no overlap.

The circle is also required to lie inside the rectangle, so its center must lie outside a strip $1/2$ wide around the edge, and hence inside an inner $19 \times 24$ rectangle, area $456$. The total area of ovals is less, so they cannot cover it completely and it must be possible to place a circle as required.