Inside a square, circles are placed, either disjoint or externally tangent to each other. If the side of the square measures $1 \mathrm{~cm}$, is it possible to place so many of these circles that the sum of their radii, in centimeters, is greater than 2008?
The answer to this problem is affirmative: it is possible to place a certain number of non-overlapping circles inside a square with a side of 1 centimeter, such that the sum of the radii of these circles is greater than 2008 centimeters.

To display such a configuration, we draw lines parallel to the sides of the square, dividing it into \( n^2 \) smaller squares; each of these smaller squares has a side length of \( 1 / n \). Inside each of these smaller squares, we draw an inscribed circle with a radius of \( 1 / 2n \). In the particular case \( n=4 \), this construction is shown in the figure.

![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-087.jpg?height=323&width=329&top_left_y=969&top_left_x=652)

\[
\left\{\begin{array}{l}
n^{2}=4^{2}=16 \text { circles } \\
\text { sides of the smaller squares }=\frac{1}{4} \\
\text { radius of the circles }=\frac{1}{8} \\
\text { sum of the radii }=16 \times \frac{1}{8}=\frac{4}{2}=2
\end{array}\right.
\]

Thus, the sum of the radii of the \( n^2 \) circles is equal to \( n^2 \times 1 / 2n = n / 2 \). Since we are interested in the case where this sum is greater than 2008, we must have \( n / 2 > 2008 \), or \( n > 4016 \). Therefore, by dividing the square into \( 4017^2 \) smaller squares (or more), the sum of the radii of the circles inscribed in the smaller squares will be greater than 2008.