Find all $10$-digit whole numbers $N$, such that first $10$ digits of $N^2$ coincide with the digits of $N$ (in the same order).
1. Let \( N = 10^9 K + X \) where \( 0 \le X \le 10^9 - 1 \). This representation helps us understand the structure of \( N \) in terms of its digits.
2. Given \( 10^9 K \le 10^9 K + X < 10^9 (K+1) \), we know that \( 10^{18} K^2 \le N^2 < 10^{18} (K+1)^2 \). This inequality helps us bound \( N^2 \).
3. Since \( K \le 9 \), we have \( (K+1)^2 \le (9+1)^2 = 100 \). Therefore, \( 10^{18} \le 10^{18} K^2 \le N^2 < 10^{18} (K+1)^2 \le 10^{20} \). This means that \( N^2 \) is either 19 digits or 20 digits long.

**Case 1: \( N^2 \) is 19 digits long.**
4. We are given that the first 10 digits of \( N^2 \) coincide with the digits of \( N \). Since \( N^2 \) is 19 digits long, we get \( 10^9 N \le N^2 < 10^9 (N+1) \).
5. Therefore, \( 10^9 \le N < 10^9 + \frac{10^9}{N} \). This implies that \( N = 10^9 \).

**Case 2: \( N^2 \) is 20 digits long.**
6. Given that the first 10 digits of \( N^2 \) coincide with the digits of \( N \), and since \( N^2 \) has 20 digits, we have \( 10^{10} N \le N^2 < 10^{10} (N+1) \).
7. This means \( 10^{10} \le N \le 10^{10} + \frac{10^{10}}{N} < 10^{10} + \frac{10^{10}}{10^9} = 10^{10} + 10 \).
8. This gives us possible solutions \( 1000000000, 1000000001, \ldots, 1000000009 \). However, none of these have perfect squares that are 20 digits long (because all of them are less than \( 1000000010 \), which has square \( 1000000020000000100 \)). Therefore, there are no solutions in this case.

This leaves us with the lone solution \( N = 10^9 = 1000000000 \).

The final answer is \( \boxed{1000000000} \)