Knapsack problem 0/1 with even number of items
For the base case, the value is often decided manually based on the context of the problem. What should be the base value?

It is easy to see that $DP_{even}[0,0]$ should be $0$, and $DP_{even}[0,j]$ for $j&gt;0$ should be $-\infty$. This is because the highest value you can get with considering none of the items and cost nothing is trivially 0, but the case where you consider none them but still cost something is impossible, hence we can safely assign it $-\infty$.

Now let's look at $DP_{odd}$. You might think that $DP_{odd}[0,0]$ is $0$, but following your definition, $DP_{odd}[0,0]$ means the highest value you can get with considering none of the items, and take an odd number of items and costs nothing. Notice the contradiction in this statement? You don't consider any items, then obviously you couldn't have taken an odd number of items! The same argument holds for all $DP_{odd}[0,j]$ for $j\geq 0$. Hence, they are all assigned $-\infty$.

To summarize, the base case is very similar. Both $DP_{even}[0,j]$ and $DP_{odd}[0,j]$ are mostly $-\infty$, except $DP_{even}[0,0]$ which is $0$. Now, you can safely run your recurrence relations on top of the base cases.