1. Given that the set $M$ is a subset of $\{1,2, \cdots, 2011\}$, and the sum of any four elements in $M$ cannot be divisible by 3. Then $|M|_{\text {max }}=$ $\qquad$
- 1. 672.

Consider the set $A=\{3,6,9, \cdots, 2010\}$,
$$
\begin{array}{l}
B=\{1,4,7, \cdots, 2011\}, \\
C=\{2,5,8, \cdots, 2009\} .
\end{array}
$$

If $M \cap A \neq \varnothing$, then $|M \cap B|<3,|M \cap C|<3,|M \cap A|<4$. Therefore, $|M|<10$.
Now assume $M \cap A=\varnothing$.
If $|M \cap B| \geqslant 2$, then $|M \cap C|<2$.
$$
\begin{array}{l}
\text { Hence }|M|=|M \cap B|+|M \cap C| \\
\leqslant|B|+1=672 .
\end{array}
$$

If $|M \cap B| \leqslant 1$, then $|M| \leqslant|C|+1=671$. Therefore, $|M|_{\max }=672$.
For example, take $M=B \cup\{2\}$.