Denote the value of the given expression by $k$. We know that $2024 = 2^3 \cdot 11 \cdot 23$ and $CC = C \cdot 11$. In a perfect square, all prime exponents are even. Thus one of the numbers $AB$, $BA$, $CC$ must be divisible by $23$. This cannot be $CC$, so let it be $AB$ without loss of generality (the other option is symmetrical). The two-digit multiples of $23$ are $23$, $46$, $69$ and $92$.

* If $AB = 23$, then $BA = 32 = 2^5$, so $k = 2^8 \cdot 11^2 \cdot 23^2 \cdot C$. Then $k$ is a perfect square iff $C$ is a perfect square, meaning $C = 1$, $C = 4$ or $C = 9$. The sum $A+B+C$ is $6$, $9$ or $14$ respectively.

* If $AB = 46$, then $BA = 64 = 2^6$, so $k = 2^{10} \cdot 11^2 \cdot 23^2 \cdot C$. Then $k$ is a perfect square iff $C$ is a perfect square, meaning $C = 1$ or $C = 9$ (the digit $4$ is already in use). The sum $A+B+C$ is $11$ or $19$ respectively.

* If $AB = 69$, then $BA = 96 = 2^5 \cdot 3$, so $k = 2^8 \cdot 3^2 \cdot 11^2 \cdot 23^2 \cdot C$. Then $k$ is a perfect square iff $C$ is a perfect square, meaning $C = 1$ or $C = 4$ (the digit $9$ is already in use). The sum $A+B+C$ is $16$ or $19$ respectively.

* If $AB = 92$, then $BA = 29$, so $k = 2^5 \cdot 11^2 \cdot 23^2 \cdot 29 \cdot C$. This cannot be a perfect square, as $C$ cannot be divisible by $29$.

The sum $A+B+C$ can thus obtain $6$ different values ($19$ was present in two different cases).