10. Now, a cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with a side length of 101 is to be divided into two parts, with the following requirements: (1) The cutting plane intersects the edges $A A_{1}, B B_{1}, C C_{1}, D D_{1}$ at points $P, Q, R, S$ (which can coincide with the vertices); (2) The lengths of segments $A P, B Q, C R, D S$ are all non-negative integers, and each set of values for segments $A P, B Q, C R, D S$ corresponds to a unique way of dividing, then there are $\qquad$ different ways of dividing. (Answer with a number)
Let $A P=a, B Q=b, C R=c, D S=d$. According to the problem, $a+c=b+d$. Considering the number of non-negative integer solutions for $a+c=k(k \in \mathbf{N}, k \leqslant 101)$ is $C_{k+2-1}^{1}=k+1$, and the number of non-negative integer solutions for $a+c=k(k \in \mathbf{N}, 102 \leqslant k \leqslant 202)$ is $C_{k+2-1}^{1}-2 C_{k-102+2-1}^{1}=203-k$. Since the cross-section $A B C D$ when $k=0$ and the cross-section $A_{1} B_{1} C_{1} D_{1}$ when $k=202$ do not meet the requirements, the number of non-negative integer solutions $(a, b, c, d)$ satisfying $a+c=b+d=k$ is
\[
\left\{\begin{array}{l}
(k+1)^{2}, 1 \leqslant k \leqslant 101, \\
(203-k)^{2}, 102 \leqslant k \leqslant 201 .
\end{array}\right.
\]
Thus, the number of different partitioning methods is
\[
\sum_{k=1}^{101}(k+1)^{2}+\sum_{k=102}^{201}(203-k)^{2}=\frac{101 \times 102 \times 203}{3}+102^{2}-2=707504.
\]