Problem 1195

1. First, we define the sequence $T_n$ as given:
$$T_1 = 2$$
$$T_2 = 2^{T_1} = 2^2 = 4$$
$$T_3 = 2^{T_2} = 2^4 = 16$$
$$T_4 = 2^{T_3} = 2^{16}$$

2. For $n \geq 4$, $T_n = 2^{T_{n-1}}$ becomes extremely large. We need to consider these terms modulo 255. Notice that:
$$2^{16} \equiv 1 \pmod{255}$$
This is because $2^{16} = 65536$ and $65536 \mod 255 = 1$.

3. Since $T_4 = 2^{16} \equiv 1 \pmod{255}$, for $n \geq 4$, $T_n \equiv 1 \pmod{255}$.

4. Now, we need to sum the first 256 terms of the sequence modulo 255:
$$T_1 + T_2 + T_3 + T_4 + \cdots + T_{256}$$

5. We know:
$$T_1 = 2, \quad T_2 = 4, \quad T_3 = 16$$
and for $n \geq 4$, $T_n \equiv 1 \pmod{255}$.

6. Therefore, the sum can be broken down as:
$$T_1 + T_2 + T_3 + \sum_{n=4}^{256} T_n$$
$$= 2 + 4 + 16 + \sum_{n=4}^{256} 1$$

7. The number of terms from $T_4$ to $T_{256}$ is $256 - 4 + 1 = 253$.

8. Thus, the sum is:
$$2 + 4 + 16 + 253 \times 1$$
$$= 2 + 4 + 16 + 253$$
$$= 275$$

9. Finally, we find the remainder when 275 is divided by 255:
$$275 \mod 255 = 20$$

The final answer is $\boxed{20}$.