10. Satisfy $0 \leqslant k_{i} \leqslant 20(i=1,2,3,4)$, and $k_{1}+k_{3}=k_{2}+k_{4}$ of the ordered integer tuples $\left(k_{1}, k_{2}, k_{3}, k_{4}\right)$ the number is $\qquad$ .
10. 6181.

For $0 \leqslant m \leqslant 20$, the non-negative integer solutions satisfying $x+y=m$ and $0 \leqslant x, y \leqslant 20$ are
$$
(x, y)=(j, m-j)(0 \leqslant j \leqslant m),
$$

there are $m+1$ solutions;
When $20<m \leqslant 40$, the non-negative integer solutions satisfying $x+y=m$ and $0 \leqslant x, y \leqslant 20$ are
$$
(x, y)=(j, m-j)(m-20 \leqslant j \leqslant 20),
$$

there are $40-m+1$ solutions.
Therefore, the number of solutions satisfying $k_{1}+k_{3}=k_{2}+k_{4}$ is
$$
\begin{array}{l}
\sum_{m=0}^{20}(m+1)^{2}+\sum_{m=21}^{40}(40-m+1)^{2} \\
=2 \times \frac{1}{6} \times 20 \times 21 \times 41+21^{2}=6181 .
\end{array}
$$