Let $ABCD$ be a square with side $4$. Find, with proof, the biggest $k$ such that no matter how we place $k$ points into $ABCD$, such that they are on the interior but not on the sides, we always have a square with sidr length $1$, which is inside the square $ABCD$, such that it contains no points in its interior(they can be on the sides).
1. **Divide the Square:**
   Consider the square \(ABCD\) with side length \(4\). Divide this square into \(16\) smaller squares, each with side length \(1\). This can be done by drawing lines parallel to the sides of \(ABCD\) at intervals of \(1\).

2. **Pigeonhole Principle:**
   According to the pigeonhole principle, if we place \(16\) points into these \(16\) smaller squares, at least one of these squares will contain at least one point. However, we need to ensure that there is at least one square with no points in its interior.

3. **Placing Points:**
   If we place \(15\) points in the \(16\) smaller squares, by the pigeonhole principle, at least one of these squares will be empty. This empty square will be a \(1 \times 1\) square with no points in its interior.

4. **Configuration for \(16\) Points:**
   To show that \(k = 16\) is not possible, consider placing a point at the center of each of the \(16\) smaller squares. The coordinates of these points will be \((0.5, 0.5), (0.5, 1.5), \ldots, (3.5, 3.5)\).

5. **Dilating the Points:**
   Now, dilate these points from the center of the big square with a ratio \(\alpha\) such that \(\frac{2}{3} < \alpha < \frac{\sqrt{2}}{2}\). This dilation will move the points slightly but keep them within the boundaries of the smaller squares.

6. **No Empty \(1 \times 1\) Square:**
   After dilation, it is not hard to see that there will be no \(1 \times 1\) square (parallel or otherwise) that does not contain any of these points in its interior. This is because the points are distributed in such a way that they cover all possible \(1 \times 1\) squares.

Conclusion:
Thus, the largest \(k\) such that no matter how we place \(k\) points into \(ABCD\), there is always a \(1 \times 1\) square with no points in its interior is \(15\).

The final answer is \( \boxed{ 15 } \).