Suppose that we do not return balls to the box. The plain probability that the $i$-th ball selected is white is $\frac{k}{N}$, exactly like in the case of returning the ball to the box.
One way of seeing this is to number the balls, white from $1$ to $k$, black from $k+1$ to $N$. Imagine that we draw the balls, one at a time, until all the balls are gone. All permutations of the labels are equally likely, and the fraction of these permutations for which the $i$-th ball drawn is white is $\dfrac{k}{N}$. (It can take a while until this fact becomes "obvious"!)
But in drawing one at a time, there are other probabilities that we can compute, for example the probability that the $3$rd ball drawn is white given that the first two were white. This conditional probability is not $\dfrac{k}{N}$, it is $\dfrac{k-2}{N-2}$ (except in trivial cases, like $k=1$).
Suppose again that we draw the balls, one at a time, until they are all gone. The conditional probability that the $3$rd ball drawn is white, given that the last two balls drawn (of the $N$) are white is also $\dfrac{k-2}{N-2}$. So it is not the temporal order of the drawing that matters in evaluating the conditional probability.
When we calculate a conditional probability, it is not the act of receiving information that matters, it is the act of using the information to calculate a conditional probability, that is, to restrict the sample space.
When one does replacement, conditional and unconditional probabilities are the same, since the sample space is unchanged.
First, the number of balls in the box decrease by one at each step. Suppose $B(t)$ and $W(t)$ are the number of black and white balls present after $t$ steps. Since we start with $B(0)+W(0)=2731$ and $B(t+1)+W(t+1)=B(t)+W(t)-1$, we have that $B(2730)+W(2730)=1$. At this point we can no longer continue the process. So at this end, there is either a white ball or a black ball left.
Notice that the number of white balls present at any time is even. For instance, suppose we have just done step $t$. If we choose two black balls, then $B(t+1)=B(t)-1$ and $W(t+1)=W(t)$. If we choose a white ball and a black ball, then $B(t+1)=B(t)-1$ and $W(t+1)=W(t)$. If we choose two white balls, then $B(t+1)=B(t)+1$ and $W(t+1)=W(t)-2$. Necessarily, since $W(0)$ is even, it stays even throughout the process.
Hence $W(2730)=0$ and $B(2730)=1$. Even though what happens throughout the process is random, by parity, the process must end with only one black ball left.
Best Answer
You received two nice answers already, but let me add one that might be experienced as more intuitive.
If the ball has white color then the probability on drawing a red ball is $\frac12$.
If the ball has red color then the probability on drawing a red ball is $1$.
So comparing both situations you could say that the odds are $\frac12:1$.
We can also express that as $\frac13:\frac23$.
Here $\frac13+\frac23=1$ so that the two numbers can be interpreted as probabilities, and we can conclude that the first situation corresponds with probability $\frac13$.