8. For the $k$-element subset $T$ of the set $\{1,2, \cdots, 242\}$, such that the sum of any two elements of $T$ (which can be the same) is not a cube of 3. Then the maximum value of $k$ is $\qquad$
8. 121 .

Notice that, $243=3^{5}$.
Therefore, in each of the pairs $\{1,242\},\{2,241\}, \cdots,\{121,122\}$, at most one number belongs to the set $T$.
Thus, $k \leqslant 121$.
Now take $T=\{122,123, \cdots, 242\}$. Then for any two elements $a, b$ in the set $T$, we have
$$
\begin{array}{l}
3^{5}=243<122+122 \leqslant a+b \\
\leqslant 242+242<3^{6} .
\end{array}
$$

Hence, the sum of any two elements in the set $T$ is not a power of 3.
In conclusion, the maximum value of $k$ is 121.