21.3.15 **Let $L_{1}, L_{2}, L_{3}, L_{4}$ be the four legs of a square table, each with a height of a positive integer $n$. How many ordered quadruples of non-negative integers $\left(k_{1}, k_{2}, k_{3}, k_{4}\right)$ are there such that after cutting off a segment of length $k_{i}$ from each leg $L_{i}$ (starting from the ground), $i=1,2,3,4$, the table remains stable? Here, a table is considered stable if and only if all four legs can be placed on the ground simultaneously.**
Parse using spatial coordinates, let's assume the coordinates of the four feet are $(1,0,0),(0,1,0)$, $(-1,0,0),(0,-1,0)$. Then the problem turns into finding the number of non-negative integer tuples $\left(k_{1}, k_{2}, k_{3}, k_{4}\right)$ such that the points $F_{1}\left(1,0, k_{1}\right), F_{2}\left(0,1, k_{2}\right), F_{3}\left(-1,0, k_{3}\right)$, and $F_{4}\left(0,-1, k_{4}\right)$ are coplanar, where $0 \leqslant k_{i} \leqslant n, i=1,2,3,4$.

The points $F_{1}, F_{2}, F_{3}, F_{4}$ are coplanar if and only if the line $F_{1} F_{3}$ intersects with $F_{2} F_{4}$, which is equivalent to the midpoint of $F_{1} F_{3}$ coinciding with the midpoint of $F_{2} F_{4}$ (in fact, if $F_{1}, F_{2}, F_{3}, F_{4}$ are coplanar, then $F_{1} F_{2} F_{3} F_{4}$ forms a parallelogram). Thus, the problem turns into finding the number of non-negative integer tuples $\left(k_{1}, k_{2}, k_{3}, k_{4}\right)$ that satisfy: $\frac{k_{1}+k_{3}}{2}=\frac{k_{2}+k_{4}}{2}$.

Notice that, when $m \leqslant n$, the number of non-negative integer solutions $(x, y)=(j, m-j), 0 \leqslant j \leqslant m$, is $m+1$; and when $n \leqslant m \leqslant 2 n$, the solutions are $(x, y)=(j, m-j), m-n \leqslant j \leqslant n$, with a total of $2 n-m+1$ solutions.
Therefore, the number of non-negative integer tuples that satisfy $\frac{1}{2}\left(k_{1}+k_{3}\right)=\frac{1}{2}\left(k_{2}+k_{4}\right)$ is
$$
\begin{aligned}
& 1^{2}+2^{2}+\cdots+n^{2}+(n+1)^{2}+n^{2}+\cdots+1^{2} \\
= & 2 \times \frac{1}{6} n(n+1)(2 n+1)+(n+1)^{2} \\
= & \frac{1}{3}(n+1)\left(2 n^{2}+4 n+3\right) .
\end{aligned}
$$

In summary, the answer is $\frac{1}{3}(n+1)\left(2 n^{2}+4 n+3\right)$.