Since $(a+b)^2 = f(n) = n = 100a + b$, we have
$$(a+b)(a+b-1) = 99a.$$
By Euclid's lemma, one of the factors in the LHS is divisible by 11. Also, note that the two factors are coprime, so one of them is not divisible by 3, and thus the other must be divisible by 9.
Hence $a+b \equiv 0 \text{ or } 1 \pmod{9}$ and $a+b \equiv 0 \text{ or } 1 \pmod{11}$, so by the Chinese Remainder Theorem we have the following possibilities for $a+b \pmod{99}$:

$$a+b \equiv 0 \pmod{9},\ a+b \equiv 0 \pmod{11} \Rightarrow a+b \equiv 0 \pmod{99},$$
$$a+b \equiv 1 \pmod{9},\ a+b \equiv 0 \pmod{11} \Rightarrow a+b \equiv 55 \pmod{99},$$
$$a+b \equiv 0 \pmod{9},\ a+b \equiv 1 \pmod{11} \Rightarrow a+b \equiv 45 \pmod{99}, \text{ or}$$

$a+b \equiv 1 \pmod{9}$, $a+b \equiv 1 \pmod{11} \Rightarrow a+b \equiv 1 \pmod{99}$.
But $10^3 \le n < 10^4$ and $n = f(n) = (a+b)^2$ implies $32 \le a+b \le 99$. Hence the only possible values of $a+b$ are 45, 55, 99, so the only possible values of $n$ are $45^2 = 2025$, $55^2 = 3025$, and $99^2 = 9801$. It is easy to check that $f(n) = n$ in each of these three cases, so the complete set of solutions is $n = 2025, 3025, 9801$.