<s>[INST] If two consecutive numbers are removed from the series $1+2+3+\ldots+n$ the average becomes $99/4$. Find the two numbers. The initial average will be $\frac{n+1}{2}$. If the two numbers are $k$ and $k+1$ then the new average will be $\frac{n(n+1)/2-(2k+1)}{n-2}$. I couldn't figure further even though I got the relation between $n$ and $k$ in many different ways.
If the question is not clear, here is an example to explain it.
If $n=10$, the initial average will be $5\cdot 5$ {$(1+2+\cdots + 10)/10$}
Now if two consecutive numbers like $2,3$ or $8,9$ are removed from this series, the new average changes, and this new average has been given to be $99/4$, however we also don't know the value of $n$, so the question seems to be pretty difficult.
 [/INST] This approach finds that the new average lies within $\pm 1$ of the original average. This significantly narrows down possibilities, and the solution can then be found easily by elimination. 

After removing the two numbers, the new average, $a$, is given by
$$\begin{align}
\frac {99}4=a&=\frac{\frac{n(n+1)}2-(2k+1)}{n-2}\\
&=\underbrace{\frac {n+1}2}_{\text{original average, $a_0$}}+
\underbrace{\frac {n-2k}{n-2}}_{\in [-1,1] \text{ for } 1\le k\le n-1}
\end{align}$$
Hence $a$ lies within $\pm 1$ of the original average $a_0$ before removal, i.e.
$$a_0-1\;\le\; a=\frac {99}4=24.75\;\le\; a_0+1$$
As $a_0=\frac {n+1}2$, it can only be either an integer or an integer and a half, hence
$24\le a_0\le 25.5$.  
$$\begin{array} {lrrrr}
\hline{a_0(n)=\frac{n+1}2} &24&24.5&25&25.5\\
n  &47 &48 &49 &\color{red}{50}\\
n-2  &45 &46 &47 &\boxed{48} \\
a-a_0(n)=\frac {n-2k}{n-2}&\frac 34&\frac 14&-\frac14&-\frac34\\
\hline
\end{array}$$
Also, the sum of the remaining numbers $\frac {99}4 (n-2)$ must be integer, so $(n-2)$ must a multiple of $4$, the only candidate for which is $48$. 
Hence we conclude that $\color{red}{n=50, k=43}\qquad \blacksquare$
</s>