We start with a lemma.

Lemma 1. Let $a_1, a_2, \dots, a_n$ be $n$ even integers. Then there is a subsequence $i_1 < i_2 < \dots < i_k$ of $1, 2, \dots, n$ such that $a_{i_1} + \dots + a_{i_k}$ is divisible by $2n$.

Proof. Consider the following integers
$$S_i = \frac{1}{2}(a_1 + a_2 + \dots + a_i) \quad 1 \le i \le n$$
If there is some $i$ such that $n \mid S_i$, we are done. So we assume that $n$ divides none of the $S_i$'s and since there are $n$ numbers there exist $i < j$ such that $S_i \equiv S_j \pmod n$. This implies
$$n \mid \frac{1}{2}(a_{i+1} + \dots + a_j) \Rightarrow 2n \mid a_{i+1} + \dots + a_j$$
and the proof is complete. $\square$

Now for the main problem, we have two cases.

* $n$ is an odd number. The answer is $k = 2n$. Firstly, note that for $k < 2n$ if we set $a_1 = a_2 = \dots = a_k = 1$, we cannot choose an even number of $a_i$'s with sum divisible by $n$. On the other hand, if $a_1, a_2, \dots, a_{2n}$ are $2n$ integers, then we define $S_i = a_1 + \dots + a_{2i}$ for $i = 1, 2, \dots, n$. By arguments same as the proof of the lemma, there is some $i$ such that $n \mid S_i$ or there exists $i < j$ such that $n \mid S_j - S_i$. In both cases we have found an even number of $a_i$'s having sum divisible by $n$.

* $n = 2m$ is an even number. We claim that $k = n + 1 = 2m + 1$ is the answer. Note that for $k \le 2m$, if we set $a_1 = a_2 = \dots = a_{k-1} = 1$ and $a_k = 0$, it is not possible to select an even number of $a_i$'s having a sum divisible by $2m$. On the other hand, suppose that $a_1, a_2, \dots, a_{2m+1}$ are $2m+1$ arbitrary integers. Assume that $b_1, b_2, \dots, b_s$ are the even numbers among $a_i$'s and $c_1, c_2, \dots, c_r$ are the odd numbers. Since $r+s = 2m+1$, exactly one of $r$ or $s$ is odd and the other is even. We suppose that $r$ is odd and $s$ is even (the other case is similar). Now look at the $m$ numbers $a_1+a_2, a_3+a_4, \dots, a_{r-2}+a_{r-1}, b_1+b_2, \dots, b_{s-1}+b_s$. Note that all these sums are even, so referring to the lemma, we can select some of them with sum divisible by $2m = n$. Since each of these numbers is sum of two members of $a_i$'s we have found an even number of $a_i$'s having a sum divisible by $n$, as desired.