Let $n=2014$. We will prove that the maximum $k$ is equal to $2n-5$. The smallest number that can be written as the sum of a red and a blue number is $1+2=3$ and the largest number is $(n-1)+n=2n-1$. Therefore, there are at most $2n-3$ numbers that can be written as the sum of a red and a blue number.
Assume that the numbers can be colored in such a way that there are $2n-3$ or $2n-4$ numbers that can be written as the sum of a red and a blue number. There is now at most one number from 3 to $2n-1$ that cannot be written as such. We will first show that we can assume without loss of generality that this number is at least $n+1$. We can create a second coloring where a number $i$ is blue if and only if $n+1-i$ was blue in the first coloring. Then a number $m$ can be written as the sum of red and blue in the second coloring if and only if $2n+2-m$ could be written as the sum of red and blue in the first coloring. Therefore, if a number less than $n+1$ could not be written as the sum of red and blue in the first coloring, then in the second coloring, precisely a number greater than $2n+2-(n+1)=n+1$ cannot be written as the sum of red and blue.
Thus, we can assume that the numbers from 3 to $n$ can all be written as the sum of red and blue. Since red and blue are interchangeable, we can also assume without loss of generality that 1 is colored blue. Since 3 can be written as the sum of red and blue and this can only be $3=1+2$, 2 must be red. Now suppose we know that 2 to $l$ are red, for some $l$ with $2 \leq l \leq n-2$. Then in all possible sums $a+b=l+2$ with $a, b \geq 2$, both numbers are colored red, but we know that we can write $l+2$ as the sum of red and blue (since $l+2 \leq n$), so it must be $1+(l+1)$. Therefore, $l+1$ is also red. By induction, we see that the numbers from 2 to $n-1$ are all red. That is $n-2=2012$ numbers. But there are only $\frac{1}{2}n=1007$ red numbers, contradiction.
We conclude that there are at least two numbers from 3 to $2n-1$ that cannot be written as the sum of a red and a blue number. We will now show that we can color the numbers in such a way that all numbers from 4 to $2n-2$ can be written as the sum of a red and a blue number, so that the maximum $k$ is equal to $2n-5$.
For this, color all even numbers except $n$ blue and also the number 1. All odd numbers except 1 are colored red, and also the number $n$. By adding 1 to an odd number (not equal to 1), we can write all even numbers from 4 to $n$ as the sum of a red and a blue number. By adding 2 to an odd number (not equal to 1), we can write all odd numbers from 5 to $n+1$ as the sum of a red and a blue number. By adding $n-1$ to an even number (not equal to $n$), we can write all odd numbers from $n+1$ to $2n-3$ as the sum of a red and a blue number. By adding $n$ to an even number (not equal to $n$), we can write all even numbers from $n+2$ to $2n-2$ as the sum of a red and a blue number. In total, we can thus write all numbers from 4 to $2n-2$ as the sum of a red and a blue number.
We conclude that the maximum $k$ is equal to $2n-5=4023$.