Analytically, I have not gotten far. First thing I notice is that $\textrm{fib}\left(n\right)$ is only even if $n \% 3 = 0$, else it is odd.

Think in binary or follow my proof if you do not believe me:

First step, $\textrm{fib}\left(3 \cdot n\right)$ is always even:

Basis:

$$n = 0 \\ \textrm{fib}\left(0\right) = 0$$

Assumption: Holds to $n$, therefore $\textrm{fib}\left(n\right)$ is even

Induction: $n \rightarrow n + 1$

\begin{align} \textrm{fib}\left(3\cdot \left(n+1\right) \right) &= \textrm{fib}\left(3\cdot n + 3\right) \\ &= \textrm{fib}\left(3 \cdot n + 2\right) + \textrm{fib}\left(3 \cdot n + 1\right) \\ &= 2 \cdot \textrm{fib}\left(2 \cdot n + 1 \right) + \textrm{fib}\left(3 \cdot n \right) \end{align}

$2 \cdot x, x \in N$ is always even, $\textrm{fib}\left(3 \cdot n\right)$ is even and the sum of even numbers is even as well. Ergo $\textrm{fib}\left(3 \cdot n\right)$ is always even!

But we still have to proof that $\textrm{fib}\left(3 \cdot n\right)$ are the ONLY even numbers!

Lets take a look at $\textrm{fib}\left(3 \cdot n + 2\right)$ which should be always odd:

Basis:

$$n = 0 \\ \textrm{fib}\left(2\right) = 1$$

Assumption: Holds to $n$, therefore $\textrm{fib}\left(n + 2\right)$ is odd

Induction: $n \rightarrow n + 1$

\begin{align} \textrm{fib}\left(3 \cdot \left(n+1\right) + 2\right) &= \textrm{fib}\left(3 \cdot n + 5\right) \\ &= \textrm{fib}\left(3n + 4\right) + \textrm{fib}\left(3 \cdot \left(n + 1\right)\right) \\ &= 2 \cdot \textrm{fib}\left(3 \cdot \left(n + 1\right)\right) + \textrm{fib}\left(3 \cdot n + 2\right) \end{align}

A sum of a odd and an even number is always odd hence $\textrm{fib}\left(3 \cdot n+2\right)$ is odd

What is left? $\textrm{fib}\left(3 \cdot n + 1\right)$!

Basis:

$$n = 0 \\ \textrm{fib}\left(1\right) = 1$$

Assumption: Holds to $n$, therefore $\textrm{fib}\left(n + 1\right)$ is odd

Induction: $n \rightarrow n + 1$

\begin{align} \textrm{fib}\left(3 \cdot \left(n+1\right) + 1\right) &= \textrm{fib}\left(3 \cdot n + 4\right) \\ &= \textrm{fib}\left(3 \cdot n + 2 \right) + \textrm{fib}\left(3 \cdot \left(n + 1\right)\right) \\ \end{align}

And again, odd + even = odd!

So the only even numbers in the Fibonacci sequence are the ones with $n \% 3 = 0$!

But that's it, no idea how to battle this problem further.

# Update

Well, after some thinking, I guess I found an analytic solution.

Lets have look which Fibonacci numbers are in the sum of all even Fibonacci numbers: (remember, every third is even)

$$\begin{array}{c|ccccccccccccc} N & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 2 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 3 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ \end{array}$$

But these could also be written like this: $$\begin{array}{c|ccccccccccccc} N & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 2 & 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 & 0 & 0 \\ 3 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 \\ \end{array}$$

This means that $$\sum_{i=0}^N f_{\textrm{even}} = \sum_{i=0}^N f_{\textrm{odd}}$$ and therefore we can also write it as: $$\sum_{i=0}^N f_{\textrm{even}} = \frac{1}{2} \sum_{i=0}^N f_i$$

We now only need to know what the sum of all Fibonacci numbers to some n is: $$\sum_{i=0}^N f_i = f_{N+2} - 1$$

Wherefore, the solution is: $$\sum_{i=0}^N f_{\textrm{even}} = \frac{f_{n+2} - 1}{2} \ \textrm{with n chosen to be the biggest value fulfilling: } f_n \leq 4000000$$