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?

Say you're removing $x$ from the set $1, 2, ldots, n$. The average of the resulting numbers will be at least $n/2$ (if you remove $n$) and at most $(n+2)/2$ (if you remove $1$.) So we have

$$ n/2 le 163/4 le (n+2)/2 $$

or, multiplying through by 4,

$$ 2n le 163 le 2n+2. $$.

So $n = 81$. You don't have to explicitly work out the number being removed, but it's $(1 + 2 ldots + 81) - (163/4) times 80$ = $(81 times 82)/2 - (163/4) times 80 = 61$.

The average of $1,2,3,ldots,n$ is the number halfway between the endpoints, $(n+1)/2,$ so the sum is $n(n+1)/2.$ Omitting $x$ from among $1,2,3,ldots,n,$ we get the sum $n(n+1)/2-x.$

Thus the average of $1,2,3,ldots,n$ must be a weighted average of $big( n(n+1)/2-xbig)/(n-1)$ and $x,$ with respective weights $(n-1)/n$ and $1/n.$ $$ frac{n-1} n left( frac{n(n+1)/2} {n-1} - frac x {n-1} right) + frac 1 ncdot x = frac{n+1} 2 $$ Therefore we have: begin{align} & frac{n-1} n cdot frac{163} 4 + frac x n = frac{n+1} 2 \[8pt] & frac{n(n+1)}{2(n-1)} - frac x {n-1} = frac{163} 4 end{align} So we get a system of two equations that is quadratic in $n$ and linear in $x.$ I'd try solving for one of those two in terms of the other and then substituting and solving the remaining equation.

This is not how you are supposed to solve it, but I feel like cheating. We have the equation $2(n^2+n-2x)=163(n-1)$, and $1leq xleq n$.

If you assume that $x=1$ then you solve for $n$ using the quadratic formula, and you obtain $79.5$.

If you assume that $x=n$ then you solve for $n$ you obtain $81.5$. Thus $n=80$ or $n=81$.

If $n=80$ then you can solve for $x$ and obtain $83=4x$, wrong. So $n=81$. Solving again yields $4n=244$, and ding ding, we have a winner.

As everyone else seems to have done this in the way I considered cheating, I should expand on the way I initially did it, which is completely different.

We have $2(n^2+n-2x)=163(n-1)$, and $1leq xleq n$. We see that $n$ is odd, so $n=2m+1$. Substituting in and cancelling the 2s yields

$$(2m+1)^2+(2m+1)-2x=163m$$ or $$4m^2+2-2x=157m.$$ Write $y=x-1$ and also, the LHS is even, so $m=2a$ is even. More substituting and removing the $2$ from both sides yields $$8a^2-y=157a.$$ The crucial point: we see that $amid y$. Since $n=2m+1=4a+1$, and $y$ is a multiple of $a$, $y=alpha a$ for $alpha$ between $1$ and $4$. Dividing through by $a$ yields $$8a-alpha=157.$$ Taking congruences modulo $8$ yields $alphaequiv 3bmod 8$, so $alpha=3$. Thus $y=3a$, so $x=3a+1$ and $n=4a+1$. We put this back into the top equation, $2(n^2+n-2x)=163(n-1)$, to obtain $a=20$, so $n=81$, $x=61$.

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 $nleq 81.5$, so $n=81$