Complexity of finding the largest induced subgraph with all even degrees
The problem is NP-complete. We'll make a series of reductions from max-cut to show this.

Problem 0 (your problem): Given a graph $G$, does G have an induced subgraph with at least k vertices, such that all vertices have even degree within the subgraph?

Problem 1: Given a graph $G$ and subset $A$ of vertices, does $G$ have an induced subgraph with at least $k$ vertices in $A$, such that all vertices have even degree within the subgraph?

Problem 2: Given a graph $G$ and subsets $A, B$ of vertices, does $G$ have an induced subgraph with at least $k$ vertices in $A$, such that all vertices in $B$ have even degree within the subgraph?

Max-cut: Given a graph $G$, does there exist a cut $C \subset V$ of size $|\delta(C)|$ at least $k$? (where $\delta(C)$ is the set of edges in $E$ with exactly one endpoint in $C$)

We'll reduce max-cut to problem 2, problem 2 to problem 1, and finally problem 1 to problem 0, proving that your problem is NP-complete.



First we reduce the max-cut problem to problem 2. Given graph $G = (V, E)$ and $k$, we'll construct $G' = (V', E')$ and $A, B$ such that $G$ has a cut of size at least $k$ if and only if $G'$ has an induced subgraph with at least $k + |E|$ vertices in $A$, where every vertex in $B$ has even degree.

For every $v_{i} \in V$ we'll have a node $v_{i}'$ in $V'$. For every edge $e_{i} = (v_{x_{i}}, v_{y_{i}}) \in E$ we'll have nodes $e'_{i, 0}, e'_{i, 1}$ in $V'$, and edges $(e'_{i, 0}, e'_{i, 1}), (e'_{i, 0}, v'_{x_{i}}), (e'_{i, 0}, v'_{y_{i}}), (e'_{i, 1}, v'_{x_{i}}), (e'_{i, 1}, v'_{y_{i}})$. We'll set $A = B = \bigcup_{i = 1}^{|E|} \{e'_{i, 0}, e'_{i, 1}\}$.



Now, note that if we have exactly one of $v'_{x_{i}}$ or $v'_{y_{i}}$ in our subgraph, we can have both of $e'_{i, 0}$ and $e'_{i, 1}$. However, due to the degree constraint, if we have both or neither of $v'_{x_{i}}, v'_{y_{i}}$, we can have at most one of $e'_{i, 0}$ and $e'_{i, 1}$.

Take a cut $C \subset V$ in $G$ of size $k = |\delta(C)|$. We'll build $C' \subset V'$ s.t. $|C' \cap A| = |E| + k$, and every vertex in $B$ has even degree in the subgraph induced on $C'$. We set $v'_{i} \in C'$ iff $v_{i} \in C$, $e'_{i, 0} \in C'$ for all $i$, and $e'_{i, 1} \in C'$ for $e_{i} \in \delta(C)$. This construction has the claimed properties.

Next take $C' \subset V'$ s.t. $|C' \cap A| = |E| + k$, and and every vertex in $B$ has even degree in the subgraph induced on $C'$. Then we'll build a cut $C \subset V$ of size at least $k$. To do this, just set again $v_{i} \in C$ iff $v'_{i} \in C'$. Now since $|C' \cap A| = |E| + k$, there are at least $k$ edges $e_{i}$ s.t. both $e'_{i, 0}$ and $e'_{i, 1}$ are in $C'$. But for those $i$ we have $e_{i} \in \delta(C)$, hence $|\delta(C)| \geq k$ as desired.



Next we reduce problem 2 to problem 1. Given graph $G = (V, E), A, B$ and $k$, we'll construct $G' = (V', E')$ and $A'$ s.t. $G'$ has an induced subgraph with at least $2k$ vertices in $A'$ where every node has even degree in the induced subgraph if and only if $G$ has an induced subgraph with at least $k$ vertices in $A$ where every node in $B$ has even degree in the induced subgraph.

To build $G'$, We'll make two copies of $G$: for every node $v_{i} \in V$ we'll have nodes $v'_{i, 0}$ and $v'_{i, 1}$ in $G'$, and for every edge $e_{i} = (v_{x_{i}}, v_{y_{i}})$ in $E$ we'll have edges $e'_{i, 0} = (v'_{x_{i}, 0}, v'_{y_{i}, 0})$ and $e'_{i, 1} = (v'_{x_{i}, 1}, v'_{y_{i}, 1})$ in $E'$. Additionally, for every $v_{i} \in B^{c}$, we'll add a new vertex $m'_{i}$, and the edges $(v'_{i, 0}, m'_{i})$ and $(v'_{i, 1}, m'_{i})$. We set $A' = \{v'_{i, 0}, v'_{i, 1} \mid v_{i} \in A\}$.

Take a subgraph $C \subset V$ such that $|C \cap A| = k$ and vertices in $B$ have even degree in the graph induced on $C$. We'll build $C'$ s.t. $|C' \cap A'| = 2k$ and every vertex has even degree in the graph induced on $C'$. To do this, set $v'_{i, 0}, v'_{i, 1} \in C'$ if and only if $v_{i} \in C$. Then, for $v_{i} \in B^{c}$ with odd degree in the graph induced on $C$, we add $m'_{i}$ to $C'$. Now every vertex has even degree in the graph induced on $C'$, since vertices in $B$ 
have the same degree as in the graph induced on $C$ (hence even degree), we made vertices in $B^{c}$ have even degree, and since $C'$ is the same in both copies of the graph, every vertex $m'_{i}$ has degree $2$ or $0$. Clearly $|C' \cap A| = 2k$.

Now take any subgraph $C' \subset V'$ s.t. $|C' \cap A'| \geq 2k$, and every vertex has even degree in the graph induced on $C'$. We'll construct $C$ s.t. $|C \cap A| \geq k$, and every vertex in $B$ has even degree in the graph induced on $C$. To do this, we first build $C_{0}, C_{1}$, where $v_{i} \in C_{0}$ iff $v'_{i, 0} \in C'$, and $v_{i} \in C_{1}$ iff $v'_{i, 1} \in C'$. Note that $|C_{0} \cap A| + |C_{1} \cap A| = |C' \cap A'| \geq 2k$, hence WLOG $|C_{0} \cap A| \geq k$, and we set $C = C_{0}$. If some vertex $v_{i}$ in $B$ had odd degree in the graph induced on $C$, then $v'_{i, 0}$ would have odd degree in the graph induced on $C'$, hence $C$ has the claimed properties.



The reduction of problem 1 to problem 0 is by far the easiest, and I won't write a detailed proof here. The idea is to add for every vertex $v_{i}$ in $A^{c}$ the vertex $v'_{i}$ and the edge $(v_{i}, v'_{i})$, and finally subtract $|A^{c}|$ from $k$. Now every maximal solution includes one of $v_{i}$ and $v'_{i}$, but cannot include both, since then $v'_{i}$ would have odd degree, hence including $v_{i}$ no longer increases the size of the graph we're inducing on.