10 In the Cartesian coordinate system, given the point set $I=\{(x, y) \mid x, y$ are integers, and $0 \leqslant x \leqslant 5,0 \leqslant y \leqslant 5\}$, then the number of different positioned squares with vertices from the set $I$ is $\qquad$.
10105 Prompt: It is easy to know that there are only two types of squares that satisfy the conditions: squares whose sides lie on lines perpendicular to the coordinate axes, called "standard squares"; and squares whose sides lie on lines not perpendicular to the coordinate axes, called "oblique squares".
(1) Among the "standard squares", there are $5^{2}$ squares with side length 1; $4^{2}$ squares with side length 2; $3^{2}$ squares with side length 3; $2^{2}$ squares with side length 4; and $1^{2}$ squares with side length 5. (In general, there are $(6-k)^{2}$ squares with side length $k$)
(2) Since the "oblique squares" with vertices in the point set $I$ are all inscribed squares of some "standard square", we only need to consider the number of inscribed squares of "standard squares".

Obviously, a "standard square" with side length 1 has no inscribed squares; a "standard square" with side length 2 has 1 inscribed square; a "standard square" with side length 3 has 2 inscribed squares; a "standard square" with side length 4 has 3 inscribed squares; and a "standard square" with side length 5 has 4 inscribed squares. (In general, a square with side length $k$ has $(k-1)$ inscribed squares)
Combining (1) and (2), we know that the number of squares that satisfy the conditions is
$$
\begin{array}{c}
5^{2}+4^{2}+3^{2}+2^{2}+1^{2}+0 \times 5^{2}+1 \times 4^{2}+2 \times 3^{2}+ \\
3 \times 2^{2}+4 \times 1^{2}=105 \text { (squares). }
\end{array}
$$