3. Several pairwise non-overlapping isosceles right triangles with leg lengths of 1 are placed on a $100 \times 100$ grid paper. It is known that the hypotenuse of any right triangle is the diagonal of some unit square; each side of a unit square is the leg of a unique right triangle. A unit square whose diagonals are not the hypotenuse of any right triangle is called a "blank square." Find the maximum number of blank squares.
3. For a general $2 n \times 2 n$ grid paper, the maximum number of empty cells is $n(n-1)$.

In fact, the $2 n \times 2 n$ grid paper is enclosed by $2 n+1$ horizontal lines and $2 n+1$ vertical lines:
$$
\begin{array}{l}
\{(x, y) \mid x=k, 0 \leqslant k \leqslant 2 n, k \in \mathbf{Z}\}, \\
\{(x, y) \mid y=k, 0 \leqslant k \leqslant 2 n, k \in \mathbf{Z}\} .
\end{array}
$$

For an isosceles right triangle, if its right-angle vertex is at the bottom-left (top-left, bottom-right, top-right) corner of its cell, the isosceles triangle is called bottom-left (top-left, bottom-right, top-right). Bottom-left (top-left) and bottom-right (top-right) triangles are collectively referred to as bottom (top) triangles.

Let $u_{k}(0 \leqslant k \leqslant 2 n)$ denote the number of bottom right triangles with one leg on the line $y=k$, and $d_{k}(0 \leqslant k \leqslant 2 n)$ denote the number of top right triangles with one leg on the line $y=k$.
$$
\begin{array}{l}
\text { Then } u_{k}+d_{k}=2 n, u_{0}=d_{2 n}=2 n, \\
u_{k}+d_{k+1}=2 n+1 .
\end{array}
$$

Solving this, we get $u_{k}=2 n-k, d_{k}=k$. There are $u_{k}$ bottom triangles and $d_{k+1}$ top triangles between the lines $y=k$ and $y=k+1$. Therefore, the number of empty cells in this row is no more than
$$
2 n-\max \left\{u_{k}, d_{k+1}\right\} \text {. }
$$

Thus, the total number of empty cells is no more than
$$
2[0+1+\cdots+(n-1)]=n(n-1) \text {. }
$$

Below is an example with $n(n-1)$ empty cells.
The set of coordinates for the right-angle vertices of bottom-left triangles:
$$
\begin{array}{l}
\{(x, y) \mid x 、 y \in \mathbf{Z}, x 、 y \geqslant 0, x+y \leqslant n-1\} \cup \\
\{(x+k-1, n+k-x) \mid x 、 k=1,2, \cdots, n\},
\end{array}
$$

The set of coordinates for the right-angle vertices of top-right triangles:
$$
\begin{array}{l}
\{(2 n-x, 2 n-y) \mid x 、 y \in \mathbf{Z}, x 、 y \geqslant 0, \\
x+y \leqslant n-1\} \cup \\
\{(x+k-1, n+k-x) \mid x 、 k=1,2, \cdots, n\},
\end{array}
$$

The set of coordinates for the right-angle vertices of top-left triangles:
$$
\begin{array}{l}
\{(x, y) \mid x 、 y \in \mathbf{Z}, x=k, n+1+k \leqslant y \leqslant \\
2 n, k=0,1, \cdots, n-1\},
\end{array}
$$

The set of coordinates for the right-angle vertices of bottom-right triangles:
$$
\begin{array}{l}
\{(2 n-x, 2 n-y) \mid x, y \in \mathbf{Z}, x=k, \\
n+1+k \leqslant y \leqslant 2 n, k=0,1, \cdots, n-1\} .
\end{array}
$$

Thus, the required result is 2450.