Define a sequence of integers by $T_1 = 2$ and for $n\ge2$, $T_n = 2^{T_{n-1}}$. Find the remainder when $T_1 + T_2 + \cdots + T_{256}$ is divided by 255.

[i]Ray Li.[/i]
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 \(\boxed{20}\).