<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] We have, that the sum of $n+1$ terms, excluding the $m$-th and $m+1$-th, is:
$$
\begin{gathered}
  S(n + 1,m) = \sum\limits_{1\, \leqslant \,k\, \leqslant \,m - 1} k  + \sum\limits_{m + 2\, \leqslant \,k\, \leqslant \,n + 1} k  = \sum\limits_{1\, \leqslant \,k\, \leqslant \,m - 1} k  + \sum\limits_{1\, \leqslant \,k\, \leqslant \,n - m} {\left( {k + m + 1} \right)}  =  \hfill \\
   = \left( \begin{gathered}
  m \\ 
  2 \\ 
\end{gathered}  \right) + \left( {m + 1} \right)\left( {n - m} \right) + \left( \begin{gathered}
  n + 1 - m \\ 
  2 \\ 
\end{gathered}  \right) =  \hfill \\
   = \frac{1}
{2}m\left( {m - 1} \right) + \left( {m + 1} \right)\left( {n - m} \right) + \frac{1}
{2}\left( {n + 1 - m} \right)\left( {n - m} \right) =  \hfill \\
   = \frac{1}
{2}\left( {m\left( {n - 1} \right) + \left( {n + 3} \right)\left( {n - m} \right)} \right) =  \hfill \\
   = \frac{1}
{2}\left( {n\left( {n - 1} \right) + 4\left( {n - m} \right)} \right) = \frac{{n\left( {n + 3} \right)}}
{2} - 2m \hfill \\ 
\end{gathered} 
$$
So we shall have:
$$
\begin{gathered}
  \frac{{S(n + 1,m)}}
{{n - 1}} = \frac{{99}}
{4}\quad  \Rightarrow \quad \left\{ \begin{gathered}
  n - 1 = 4\,q \hfill \\
  S(n + 1,m) = \frac{{n\left( {n + 3} \right)}}
{2} - 2m = 99\;q \hfill \\ 
\end{gathered}  \right.\quad  \Rightarrow  \hfill \\
   \Rightarrow \quad \left\{ \begin{gathered}
  1 \leqslant m \leqslant n = 4\,q + 1 \hfill \\
  n\left( {n + 3} \right) = 198\;q + 4m \hfill \\ 
\end{gathered}  \right. \hfill \\ 
\end{gathered} 
$$
The last gives:
$$
\begin{gathered}
  4 \leqslant 4\left( {4q + 1} \right)\left( {q + 1} \right) - 198\;q = 4m \leqslant 4\left( {4\,q + 1} \right) \hfill \\
  0 \leqslant \left( {4q + 1} \right)\left( {q + 1} \right) - \frac{{198}}
{4}\;q - 1 \leqslant 4\,q \hfill \\
  0 \leqslant q^{\,2}  - \frac{{178}}
{{16}}\;q \leqslant \,q \hfill \\
  0 \leqslant q - \frac{{178}}
{{16}} \leqslant \,1 \hfill \\ 
\end{gathered} 
$$
i.e.
$$
\frac{{178}}
{{16}} \leqslant q \leqslant \,\frac{{194}}
{{16}}\quad  \Rightarrow \quad \left\lceil {\frac{{178}}
{{16}}} \right\rceil  \leqslant q \leqslant \,\left\lfloor {10 + \frac{{34}}
{{16}}} \right\rfloor \quad  \Rightarrow \quad 12 \leqslant q \leqslant 12
$$
In conclusion, so we have:
$$
\left\{ \begin{gathered}
  q = 12 \hfill \\
  n = 4\,q + 1 = 49 \hfill \\
  m = \frac{1}
{4}\left( {n\left( {n + 3} \right) - 198\;q} \right) = 43 \hfill \\ 
\end{gathered}  \right.
$$
which in fact gives:
$$
\frac{{S(n + 1,m)}}
{{n - 1}} = \frac{{\frac{{n\left( {n + 3} \right)}}
{2} - 2m}}
{{n - 1}} = \frac{{\frac{{49 \cdot 52}}
{2} - 86}}
{{48}} = \frac{{1188}}
{{48}} = \frac{{99}}
{4}
$$
</s>