<s>[INST] One number is removed from the set of integers from $1$ to $n.$ The average of the remaining numbers is $163/4.$ Which integer was removed? 
One number is removed from the set of integers from $1$ to $n.$ The average of the remaining numbers is $\dfrac{163}4$.  Which integer was removed?

Source. British Mathematical Olympiad 2010/11, Round 1, Problem 1
I was hoping if someone could spot the flaw in my working for this question.
Attempt.
I began by letting the integer that was removed be $x$.
Then: $$\frac{1 + 2 + \cdots + (x-1) + (x+1) +\cdots + n} {n-1} = \frac{163}{4}$$
There is two arithmetic sums in the denominator, the first from 1 to $x$ and the second from $x+1$ to $n$.
These are equal to $\frac{x(x-1)}{2}$ and $\frac{(n-x)(n+x+1)}{2}$, and subbing in to first equation this gives:
$$\frac{x(x-1) + (n-x)(n+x+1)}{2(n-1)} = \frac{163}{4}$$
which reduces to:
$$\frac{n^2 + n - 2x}{2(n-1)} = \frac {163}{4}$$
And then:
$$2(n^2 + n -2x) = 163(n-1)$$
At first I thought you could consider factors, as 163 was prime then:
$n-1 = 2$ giving $n = 3$ and $n^2 + n - 2x = 163$, which using $n=3$ gives $x= -75.5$ which isn't our positive integer.
I then tried considering a quadratic in $n$ and using the discriminant but again that just looked to give a negative value of $x.$ I would be grateful for any help
 [/INST] We know that $n$ is odd. Notice that
$$ n^2 - n \leq n^2 + n -2x \leq n^2 +n - 2 $$
$$ \implies n^2 - n \leq \frac{163}{2}(n-1) \leq n^2 +n - 2  $$
which gives us $n \geq 79.5$ and $n\leq 81.5$, so $n=81$
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