Let $x$ be the number formed from the first three digits of $n$. Then $n$ is equal to $1000x + (x + 4) = 1001x + 4$ which should be the square $y^2$ of an integer $y$. The equation $y^2 = 1001x + 4$ can be written as $(y - 2)(y + 2) = 7 \cdot 11 \cdot 13 \cdot x$. Since $n < 10^6$, we have $y < 10^3$ and so $y$ is determined by its remainder on division by $1001 = 7 \cdot 11 \cdot 13$, which, by the Chinese Remainder Theorem, is determined by the three remainders $y \pmod{7}$, $y \pmod{11}$ and $y \pmod{13}$.

From the equation above we see that $y \equiv \pm 2 \pmod{7}$, $y \equiv \pm 2 \pmod{11}$ and $y \equiv \pm 2 \pmod{13}$. To find the eight possibilities for $y \pmod{1001}$ corresponding to the different choices of signs, we first complete the table

and then multiply each line by $-1$ and replace negative residues modulo 1001 by their corresponding smallest non-negative remainder.
Here is how the numbers in the fourth column of the table can be obtained. First we find numbers $m_1, m_2$ and $m_3$ such that
$$\text{l@ 2em l m 1 1 7 11 13 m 1 m 2 1 11 7 13 m 2 m 3 1 13 7 11 m 3.}$$
To do so, we just have to solve congruences like $11 \cdot 13 \cdot k \equiv 1 \pmod{7}$. The solutions of smallest absolute value are $m_1 = -286$, $m_2 = 364$ and $m_3 = -77$. If the numbers in the first three columns of our table are $a_1, a_2, a_3$, then the number in the fourth column is congruent to $a_1m_1 + a_2m_2 + a_3m_3$. For the first two rows, we obtain the numbers given in the table whereas for rows three and four we first get $-1146$ and $-1454$, respectively, and then have to add 2002. The remaining four solutions are
$$\text{l@ 2em l 999 -2 1001 691 -310 1001 453 -548 1001 145 -856 1001 .}$$
Because $y^2 = n$ and $n$ is a six-digit number, we can rule out that $y$ is one of the numbers 2, 145 or 310 as the squares of these numbers have less than six digits. When $y = 999$ we get $y^2 = 999^2 = 998001$ and this does not give a solution to the problem. This leaves us with four solutions:
$$\begin{aligned} n &= 453^2 = 205209 \\ n &= 548^2 = 300304 \\ n &= 691^2 = 477481 \\ n &= 856^2 = 732736 \end{aligned}$$
which clearly satisfy the requirements of the problem.