A child lines up $2020^2$ pieces of bricks in a row, and then remove bricks whose positions are square numbers (i.e. the 1st, 4th, 9th, 16th, ... bricks). Then he lines up the remaining bricks again and remove those that are in a 'square position'. This process is repeated until the number of bricks remaining drops below $250$. How many bricks remain in the end?
1. **Initial Setup:**
   The child starts with \(2020^2\) bricks. The first step is to remove bricks whose positions are square numbers. The number of square numbers up to \(2020^2\) is \(2020\) (since \(2020^2\) is the largest square number less than or equal to \(2020^2\)).

   Therefore, after the first removal, the number of remaining bricks is:
   \[
   2020^2 - 2020
   \]

2. **General Case:**
   Let's denote the number of remaining bricks after the \(k\)-th removal as \(t_k\). Initially, \(t_0 = 2020^2\).

   After the first removal:
   \[
   t_1 = 2020^2 - 2020
   \]

   We can generalize this process. If \(t_k = n^2 - n\), then the number of remaining bricks after the next removal is:
   \[
   t_{k+1} = t_k - \lfloor \sqrt{t_k} \rfloor
   \]

3. **Iterative Process:**
   We need to show that if \(t_k = n^2 - n\), then \(t_{k+1} = (n-1)^2 - (n-1)\).

   Proof:
   \[
   t_k = n^2 - n
   \]
   \[
   \lfloor \sqrt{t_k} \rfloor = \lfloor \sqrt{n^2 - n} \rfloor = n - 1
   \]
   Therefore,
   \[
   t_{k+1} = t_k - \lfloor \sqrt{t_k} \rfloor = n^2 - n - (n - 1) = n^2 - 2n + 1 = (n-1)^2 - (n-1)
   \]

4. **Iteration Until Below 250 Bricks:**
   We start with \(n = 2020\) and iterate until the number of bricks is less than 250.

   \[
   t_1 = 2020^2 - 2020
   \]
   \[
   t_2 = 2019^2 - 2019
   \]
   \[
   t_3 = 2018^2 - 2018
   \]
   \[
   \vdots
   \]
   \[
   t_k = (2020 - k)^2 - (2020 - k)
   \]

   We need to find the smallest \(n\) such that:
   \[
   n^2 - n < 250
   \]

   Solving the quadratic inequality:
   \[
   n^2 - n - 250 < 0
   \]

   Using the quadratic formula:
   \[
   n = \frac{1 \pm \sqrt{1 + 4 \cdot 250}}{2} = \frac{1 \pm \sqrt{1001}}{2}
   \]

   Approximating \(\sqrt{1001} \approx 31.65\):
   \[
   n \approx \frac{1 + 31.65}{2} \approx 16.325
   \]

   Since \(n\) must be an integer, we take \(n = 16\).

   Therefore, the number of remaining bricks is:
   \[
   16^2 - 16 = 256 - 16 = 240
   \]

The final answer is \(\boxed{240}\).