4. From a square iron sheet with a side length of 10 cm, circular pieces with a diameter of 1 cm can be cut out, at most $\qquad$ pieces.


4.106.

As shown in Figure 6, it is easy to see that $\angle \mathrm{O}_{1} \mathrm{O}_{2} \mathrm{O}_{3}$ is an equilateral triangle with a side length of $1 \mathrm{~cm}$, so the height $A O_{2}$ is $\frac{\sqrt{3}}{2} \mathrm{~cm}$. Assuming we can fit $k$ rows, then
$$
\frac{\sqrt{3}}{2}(k-1)+1 \leqslant 10 \text {. }
$$

Solving this, we get $k \leqslant 1+6 \sqrt{3}$, with the integer solution being $k \leqslant 11$.
Thus, 6 rows have 10 circles, and 5 rows have 9 circles, totaling 105 circles. In fact, if we arrange the last three rows to each have 10 circles, and the 9 circles are only in 4 rows, then the total is $\frac{1}{2} \sqrt{3} \times 8+1 \times 3=4 \sqrt{3}+3<10$.
Therefore, the maximum number of circles that can be placed is $9 \times 4+10 \times 7=106$.