The problem involves determining how many ways we can cut legs of a square table such that the table remains stable. Specifically, we aim to find the number of ordered 4-tuples $(k_1, k_2, k_3, k_4)$ where the lengths of the cuts $k_i$ are non-negative integers, and the legs of the table after cutting are stable, i.e., they can all touch the floor simultaneously.

### Stability Condition

For the table to be stable, the difference in lengths between the longest leg and the shortest leg should be at most the length of the shortest leg that remains. Thus, we need to ensure the difference between the longest and shortest cut is within the original length of the leg:

$$\max(n - k_1, n - k_2, n - k_3, n - k_4) - \min(n - k_1, n - k_2, n - k_3, n - k_4) \leq n.$$

Simplifying this inequality, we have:

$$(\max(n - k_1, n - k_2, n - k_3, n - k_4)) \leq n + \min(n - k_1, n - k_2, n - k_3, n - k_4).$$

Which translates to:

$$(n - \min(k_1, k_2, k_3, k_4)) \leq n + (k_{\text{min}} - k_{\text{max}}).$$

Identifying $\min(k_1, k_2, k_3, k_4)$ as $m$ and $\max(k_1, k_2, k_3, k_4)$ as $M$:

$$M + m \leq n,$$

where $M = \max(k_1, k_2, k_3, k_4)$, and $m = \min(k_1, k_2, k_3, k_4)$.

This is equivalent to:

$$k_1 + k_2 + k_3 + k_4 \leq 4m + (n - 2m).$$

We redefine the variables to convert this into a combinatorial problem about distributing the reductions in leg lengths. Consider the space of all possible reductions in leg lengths and form a combinatorial problem:

### Combinatorial Translation

Since $k_i$ are non-negative integers and $M + m \leq n$, if we let:

$$s = n - k_1 - k_2 - k_3 - k_4,$$

then we are essentially looking for the number of solutions in non-negative integers for:

$$k_1 + k_2 + k_3 + k_4 = n - s,$$

where $0 \leq s \leq n$.

### Solving for $s$

This problem is equivalent to distributing $n - s$ into 4 categories (legs), where each category can have zero units. This boils down to a classic stars and bars combinatorial problem, where the number of distributions of $n-s$ stars into 4 bars is:

$$\binom{n-s + 3}{3}.$$

To find the total count, we sum over all valid $s$:

$$\sum_{s=0}^{n} \binom{n-s + 3}{3} = \sum_{k=3}^{n+3} \binom{k}{3} \quad \text{(variable transform)}$$

This simplifies, due to the combination identity of the sum:

This evaluates to:

$$\binom{n+3}{4}.$$

Thus, the total number of valid configurations is given by:

$$\boxed{\binom{n+3}{3}}.$$