Positive integer $n$. Question: How many ordered quadruples of non-negative integers $\left(k_{1}, k_{2}, k_{3}, k_{4}\right)$ are there such that after sawing 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 one of its 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 arrays $\left(k_{1}, k_{2}, k_{3}, k_{4}\right)$, 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 necessary and sufficient condition for points $F_{1}, F_{2}, F_{3}, F_{4}$ to be coplanar is that 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 arrays $\left(k_{1}, k_{2}, k_{3}, k_{4}\right)$ 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 arrays that satisfy $\frac{1}{n}\left(k_{1}+k_{3}\right)=\frac{1}{n}\left(k_{2}+k_{4}\right)$ is $1^{2}+2^{2}+\cdots+n^{2}+(n+1)^{2}+n^{2}+\cdots+1^{2}=2 \times \frac{1}{6} n(n+1)(2 n+1)+(n+1)^{2}=$ $\frac{1}{3}(n+1)\left(2 n^{2}+4 n+3\right)$.
In summary, the answer is $\frac{1}{2}(n+1)\left(2 n^{2}+4 n+3\right)$.