Three. (50 points) Given a square table with four legs $L_{1}, L_{2}, L_{3}, L_{4}$, each of length $n$, where $n$ is a positive integer. How many ordered quadruples $\left(k_{1}, k_{2}, k_{3}, k_{4}\right)$ exist such that after sawing off a segment of length $k_{i}$ from the end of leg $L_{i}$ $(i=1,2,3,4)$, the table remains stable? Here, $k_{1}, k_{2}, k_{3}, k_{4}$ are non-negative integers (the table is considered stable if all four legs touch the ground, and it is allowed to saw off a length of 0).
Three, Proof: By placing the table upside down, with the tabletop as the $xy$ plane, we establish a spatial rectangular coordinate system. The coordinates of the four table legs $L_{1}, L_{2}, L_{3}, L_{4}$ are $(1,0,1), (0,1,1), (-1,0,1), (0,-1,1)$, respectively.

For $i=1,2,3,4$, let the length of the table leg $L_{i}$ after sawing off a segment be $l_{i}$, then $0 \leqslant l_{i} \leqslant n$. The table is stable if and only if the four points $F_{1}(1,0,l_{1}), F_{2}(0,1,l_{2}), F_{3}(-1,0,l_{3}), F_{4}(0,-1,l_{4})$ in three-dimensional space lie on the same plane, which is equivalent to the segments $F_{1}F_{3}$ and $F_{2}F_{4}$ intersecting. That is, the midpoints of these two segments coincide. Thus, we have
$$
\left(\left(0,0, \frac{1}{2}(l_{1}+l_{3})\right)=\left(0,0, \frac{1}{2}(l_{2}+l_{4})\right) .\right.
$$

This is equivalent to, for every non-negative integer $k \leqslant 2n$, setting $S_{k}$ as the number of solutions to $x+y=k$ where $x, y$ are integers and $0 \leqslant x, y \leqslant n$. The number of stable tables is then the number of solutions (1) $N=\sum_{k=0}^{2n} S_{k}^{2}$.
When $0 \leqslant k \leqslant n$, the solutions to $x+y=k$ are the ordered pairs $(j, k-j) (0 \leqslant j \leqslant k)$, so $S_{k}=k+1$.
When $n+1 \leqslant k \leqslant 2n$, the solutions to $x+y=k$ are the ordered pairs $(j, k-j) (k-n \leqslant j \leqslant n)$, so $S_{k}=2n-k+1$. Therefore, the number of stable tables is
$$
\begin{aligned}
N & =1^{2}+2^{2}+\cdots+n^{2}+(n+1)^{2}+n^{2}+\cdots+2^{2}+1^{2}=2 \times \frac{n(n+1)(2n+1)}{6}+(n+1)^{2} \\
& =\frac{1}{3}(n+1)\left(2n^{2}+4n+3\right) .
\end{aligned}
$$