Problem of probability distribution A box contains $N$ tickets numbered $1, 2, 3,...., N$. If $m$ tickets are drawn one by one from the box without replacement, then find the mean of the sum of the numbers obtained on the tickets drawn.
I have approached the sum as below.
Let $X_i$ denote the number on the $i$th ticket drawn, where $i= 1, 2,..., m$.
The sum of the numbers obtained on the tickets drawn is $S= \sum_{i=1}^{m}X_i$
Hence, the required mean
$= E(S)
  =E(\sum_{i=1}^{m}X_i)
  =\sum_{i=1}^{m}E(X_i).$
Each $X_i$ can take the values $1, 2,...,  N$ with probability $\frac{1}{N}$.
Then $E(X_i)= \frac{N+1}{2}$
So, $E(S)= \frac{m(N+1)}{2}$
But my doubt is in the above line 'Each $X_i$ can take the values $1, 2,..., N$ with probability $\frac{1}{N}$.'
Because when the drawing is done without replacement, after each draw, the number of tickets remaining decreases by 1. So the number of values left for $X_2$ is $N-1$, not $N$.
So how can the probability be $\frac{1}{N}$?
Will anyone please explain where is the mistake?
Thanks in advance.

Your calculation is actually correct, though it is not trivial to see it.
On one hand your argument is right, after the first number is drawn, the second number ($X_2$) only has one of $N-1$ possible values. Even worse, if $X_1=1$, then the expected value of $X_2$ will be $\frac{2+3+\ldots+N}{N-1}=\frac{N^2+N-2}{2(N-1)} > \frac{N+1}2.$
OTOH, the first number could also be $X_1=N$, making the expected value for $X_2$ smaller than $\frac{N+1}2$. In the end, all of this will balance out. It should be pretty intuitive to see that if you don't consider what happens before, the value of $X_i$ can be $1,2,\ldots,N$, each with the same probability $1 \over N$. Overall, there is no higher probability to draw a $4$ than a $7$ on the third draw, for example.
The step that most people have problems with is when you wrote
$$E(\sum_{i=1}^mX_i) = \sum_{i=1}^mE(X_i)$$ 
This is correct, but most people's intuition is against this, because they say "the $X_i$ are dependend". It turns out, using the integral definition, that the equation is true even if the $X_i$ are dependend (which they of course are in our case).
