10. There are two sets of square paper pieces of the same size and the same number, one set black and one set white. Xiao Zhang first uses the white paper pieces to form a rectangle without any gaps in the middle, then uses the black paper pieces to surround the already formed white rectangle to create a larger rectangle, and then continues with white paper pieces. This process is repeated, and after Xiao Zhang has used the black paper pieces 5 times, the black and white paper pieces are exactly used up. Therefore, the black paper pieces are at least $\qquad$ pieces.
10.350.

Let the first white rectangle be $a \times b$, then the first black rectangle is $(a+2)(b+2)$, the second white rectangle is $(a+4) \times (b+4), \cdots \cdots$, and the fifth black rectangle is $(a+18)(b+18)$.

Obviously, the difference in perimeter between each black rectangle and its inner white rectangle (number of paper pieces) is 8.
Given that the number of black and white square paper pieces is equal, we have
$$
5 \times 8=(a-2)(b-2) \text {. }
$$

Solving this, we get $(a, b)=(3,42),(4,22),(6,12),(7,10)$.
Therefore, the corresponding number of black square paper pieces is
$$
\frac{1}{2}(a+18)(b+18)=630,440,360,350 \text {. }
$$

Thus, the black square paper pieces are at least 350.
Note: This problem must emphasize that "only one layer of paper pieces can be added each time (except for the first white rectangle)."

Otherwise, starting from the outermost layer, they can be taken as (black, white, black, white, white, black, white, black, white, black, black, white), with the innermost white rectangle being $1 \times 2$ (as shown in Figure 6).

Therefore, the difference in the number of black and white square paper pieces is
$$
8+8-8-8-8+8=0 .
$$

In this case, the number of black square paper pieces is $\frac{24 \times 23}{2}=276$.