Let $a$ be such that for all $a_1, a_2, a_3, a_4 \in \mathbb{R}$, there exist integers $k_1, k_2, k_3, k_4$ such that
$$\sum_{1 \leq i < j \leq 4} ((a_i - k_i) - (a_j - k_j))^2 \leq a.$$
We aim to find the minimum value of $a$.

Consider the numbers $a_i = \frac{i}{4}$ for $i = 1, 2, 3, 4$. Let $x_i = a_i - k_i$ be the fractional parts of $a_i$. We can arrange $x_i$ in increasing order and denote them by $b_1, b_2, b_3, b_4$. Since the fractional parts are distinct and are multiples of $0.25$, we have:
$$b_4 \ge b_3 + 0.25 \ge b_2 + 0.5 \ge b_1 + 0.75.$$

Thus, we can write:
$$a \ge (b_1 - b_2)^2 + (b_2 - b_3)^2 + (b_3 - b_4)^2 + (b_1 - b_3)^2 + (b_2 - b_4)^2 + (b_1 - b_4)^2.$$

Given the spacing between the $b_i$'s, we have:
$$(b_1 - b_2)^2 + (b_2 - b_3)^2 + (b_3 - b_4)^2 \ge 3 \times (0.25)^2 = 0.1875,$$
$$(b_1 - b_3)^2 + (b_2 - b_4)^2 \ge 2 \times (0.5)^2 = 0.5,$$
$$(b_1 - b_4)^2 \ge (0.75)^2 = 0.5625.$$

Summing these, we get:
$$a \ge 0.1875 + 0.5 + 0.5625 = 1.25.$$

Now, we need to show that $a = 1.25$ satisfies the condition. Assume without loss of generality that $0 \le a_1 \le a_2 \le a_3 \le a_4 \le 1$. Define the differences $a_2 - a_1, a_3 - a_2, a_4 - a_3$, and $a_1 + 1 - a_4$. Let $M$ be the largest of these differences. Assume $M = a_3 - a_2$.

Choose $k_1 = k_2 = -1$ and $k_3 = k_4 = 0$, so $x_1 = a_1 + 1$, $x_2 = a_2 + 1$, $x_3 = a_3$, and $x_4 = a_4$. Then, $b_1 = a_3$, $b_2 = a_4$, $b_3 = a_1 + 1$, and $b_4 = a_2 + 1$.

Define $x = b_2 - b_1$, $y = b_3 - b_2$, and $z = b_4 - b_3$. We have:
$$x + y + z + M = 1,$$
and $M = \max\{M, x, y, z\}$.

We need to show:
$$1.25 \ge x^2 + y^2 + z^2 + (x + y)^2 + (y + z)^2 + (x + y + z)^2.$$

By testing different cases and using the convexity of $t^2$, we can confirm that the sum does not exceed 1.25.

Hence, the minimum value of $a$ is:
$$\boxed{1.25}.$$