First, it is easy to see that $k=2010$ meets the conditions. In fact, let $A_{i}=\{i, 2011\}(i=1,2, \cdots, 2010)$, then for any $1 \leqslant i<j \leqslant 2010$, we have $A_{i} \cap A_{j}=\{2011\}$, which is not an even set, so $k=2010$ meets the conditions.

From this, we can conjecture that $k \leqslant 2010$, and we will use proof by contradiction. Assume $k \geqslant 2011$, then there exist 2011 even subsets of $M$: $A_{1}, A_{2}, \cdots, A_{2011}$, such that for any $1 \leqslant i<j \leqslant 2011$, $A_{i} \cap A_{j}$ is not an even set.

For $A \subseteq M$, define $\overrightarrow{\alpha_{A}}=\left\{a_{1}, a_{2}, \cdots, a_{2011}\right\}$, where $a_{j}=\left\{\begin{array}{l}1, \text { when } j \in A, \\ 0 \text {, when } j \notin A,\end{array}\right.$ Then, $A_{i} \cap A_{j}$ is an even set if and only if $\overrightarrow{\alpha_{A_{i}}} \cdot \overrightarrow{\alpha_{A_{j}}}$ is even. Therefore, for any $i \neq j$, we have
$$\overrightarrow{\alpha_{A_{i}}} \cdot \overrightarrow{\alpha_{A_{j}}} \equiv 1(\bmod 2)$$

For $X \subseteq M$, define $\overrightarrow{S_{X}}=\sum_{x \in X} \overrightarrow{\alpha_{A_{x}}}$, we first prove that for any $X \neq \Phi$, $\overrightarrow{S_{X}} \neq(0, 0, \cdots, 0)(\bmod 2)$.

Actually, assume $\vec{S}_{X} \equiv(0,0, \cdots, 0)$, on one hand, take $u \in X$, we have
$$0 \equiv \overrightarrow{S_{X}} \cdot \overrightarrow{\alpha_{A_{x}}}=\left(\sum_{x \in X} \overrightarrow{\alpha_{A_{x}}}\right) \cdot \overrightarrow{\alpha_{A_{x}}}=\sum_{x \in X}\left(\overrightarrow{\alpha_{A_{x}}} \cdot \overrightarrow{\alpha_{A_{x}}}\right)=\overrightarrow{\alpha_{A_{x}}} \cdot \overrightarrow{\alpha_{A_{x}}}+\sum_{x \in X|x\rangle)}\left(\overrightarrow{\alpha_{A_{x}}} \cdot \overrightarrow{\alpha_{A_{x}}}\right)$$

Since $A_{u}$ is an even set, $\overrightarrow{a_{A_{u}}} \cdot \overrightarrow{\alpha_{A_{x}}} \equiv 0$, and when $x \neq u$, by (1), $\overrightarrow{\alpha_{A_{x}}} \cdot \overrightarrow{\alpha_{A_{u}}} \equiv 1$.
So $0 \equiv \overrightarrow{\alpha_{A_{x}}} \cdot \overrightarrow{\alpha_{A_{x}}}+\sum_{x \in X \mid(u)}\left(\overrightarrow{\alpha_{A_{x}}} \cdot \overrightarrow{\alpha_{A_{x}}}\right) \equiv 0+\sum_{x \in X \backslash\{u\}} 1=|X|-1(\bmod 2)$, i.e., $|X|$ is odd.

On the other hand, note that $|M|=2011$ is even, so $X \neq M$. Take $v \notin X$, we have
$$0 \equiv \overrightarrow{S_{X}} \cdot \overrightarrow{\alpha_{A_{v}}}=\left(\sum_{x \in X} \overrightarrow{\alpha_{A_{x}}}\right) \cdot \overrightarrow{\alpha_{A_{v}}}=\sum_{x \in X}\left(\overrightarrow{\alpha_{A_{x}}} \cdot \overrightarrow{\alpha_{A_{v}}}\right)$$

Since $v \notin X$, so $x \neq v$, thus by (*), $\overrightarrow{\alpha_{A_{x}}} \cdot \overrightarrow{\alpha_{A_{v}}} \equiv 1$, so
$$0 \equiv \sum_{x \in X}\left(\overrightarrow{\alpha_{A_{x}}} \cdot \overrightarrow{\alpha_{A_{v}}}\right) \equiv \sum_{x \in X} 1=|X|$$

So $|X|$ is even, which is a contradiction.
Therefore, for any $X \neq \Phi$, $\overrightarrow{S_{X}} \neq(0,0, \cdots, 0)(\bmod 2)$.
From this, we can see that for $X \neq Y$, $\vec{S}_{X} \equiv \vec{S}_{Y}(\bmod 2)$ (actually, if $\vec{S}_{X} \equiv \overrightarrow{S_{Y}}$, let $T=(X \cup Y) \backslash(X \cap Y)$, then $\overrightarrow{S_{T}}=\overrightarrow{S_{X}}+\overrightarrow{S_{Y}}-2 \overrightarrow{S_{X \cap Y}} \equiv \overrightarrow{S_{X}}+\overrightarrow{S_{Y}} \equiv(0,0, \cdots, 0)$, which is a contradiction). Thus, when $X$ takes all subsets of $M$, we can get $2^{2011}$ different vectors $\vec{S}_{x}$ in the sense of modulo 2.

However, since $\overrightarrow{S_{X}}=\sum_{x \in X} \overrightarrow{\alpha_{A_{x}}}$, and $A_{x}$ is an even set, the sum of the components of $A_{x}$ is even, so the sum of the components of $\vec{S}_{x}$ is even, and the parity of the 2011th component of $\vec{S}_{X}$ is uniquely determined by the sum of the first 2010 components. Therefore, $\vec{S}_{X}$ has only $2^{2010}$ possible values in the sense of modulo 2, which is a contradiction, so $k \leqslant 2010$.

In conclusion, the maximum value of $k$ is 2010.
Example 8 If a set contains an even number of elements, it is called an even set. Let $M=\{1,2, \cdots, 2011\}$, if there exist $k$ even subsets of $M$: $A_{1}, A_{2}, \cdots, A_{k}$, such that for any $1 \leqslant i<j \leqslant k$, $A_{i} \cap A_{j}$ is not an even set, find the maximum value of $k$. (Original problem)