(10) The number of ordered integer tuples $\left(k_{1}, k_{2}, k_{3}, k_{4}\right)$ that satisfy $0 \leqslant k_{i} \leqslant 20, i=1,2,3,4$, and $k_{1}+k_{3}=k_{2}+k_{4}$ is $\qquad$.
(10) 6181 Prompt: When $0 \leqslant m \leqslant 20$, the non-negative integer solutions $(x, y)=(j, m-j), 0 \leqslant j \leqslant m$, satisfying $x+y=m$ and $0 \leqslant x, y \leqslant 20$ are a total of $m+1$ groups.

When $21 \leqslant m \leqslant 40$, the non-negative integer solutions $(x, y)=(j, m-j), m-20 \leqslant j \leqslant 20$, satisfying $x+y=m$ and $0 \leqslant x, y \leqslant 20$ are a total of $40-m+1$ groups.
Therefore, the solutions satisfying $k_{1}+k_{3}=k_{2}+k_{4}$ are
$$
\begin{aligned}
& \sum_{m=0}^{20}(m+1)^{2}+\sum_{m=21}^{40}(40-m+1)^{2} \\
= & 2 \times \frac{1}{6} \times 20 \times 21 \times 41+21^{2}=6181 .
\end{aligned}
$$