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.
 A: The new average will be $$\frac{\frac{n(n+1)}2-(2k+1)}{n-2}=\frac {99}4$$ Solving for $k$ gives $$k=\frac{2 n^2-97 n+194}{8} $$ which must be a positive integer lower or equal to $n$.
Solving for $n$ gives $$n=
\frac{97+\sqrt{64 k+7857}}{4} $$ which must be an integer.
Now consider the extreme cases $k=1$ and $k=n-1$; this gives very narrow bounds for $n$. From algebra, $k=1\to n=\frac{93}2$ and $k=n-1\to n=\frac{101}2$.  In the worst case, only four values of $n$ would need to be tested.
Does this help you ?
A: 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 $\color{red}{n=50, k=43}\qquad \blacksquare$
A: 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:
$$
\frac{{S(n + 1,m)}}
{{n - 1}} = \frac{{\frac{{n\left( {n + 3} \right)}}
{2} - 2m}}
{{n - 1}} = \frac{{\frac{{49 \cdot 52}}
{2} - 86}}
{{48}} = \frac{{1188}}
{{48}} = \frac{{99}}
{4}
$$
A: As in Claude Leibovici's answer, removing $k+(k+1)$ from $1+2+\cdots+n$ to leave an average of $99/4$ implies, after a bit of algebra, that
$$k={2n^2-97n+194\over8}$$
is an integer between $1$ and $n-1$.  It's easy to see that $k$ being an integer implies $n\equiv2$ mod $8$.  If we write $n=8m+2$ (with $m\gt0$, since $n=2$ is obviously not possible), we find, after a tad more algebra, that $k=16m^2-89m+1$.  The inequality constraints are thus now
$$1\le16m^2-89m+1\le8m+1$$
or
$$89\le16m\le97$$
There is clearly only one integer solution:  $m=6$, corresponding to $n=50$ and $k=43$.
A: This is a calculated guessing approach. The average of $ n $consecutive integers will be. $ (n+1)/2$. $99/4$ is a little less than$ 25$. So $n $can be taken to be$ 50$. Not$ 49,$ because two less than$ 49$, which is $47 $is not divisible by $4$. So first$ 50 $numbers sum will be$ 1275$. $99/4 * 48$ will be$ 1188.$ Here $48 $is divisible by $4$ is the reason I took $ n=50. $.  $ 1275-1188$ will be$ 87$. two consecutive numbers giving $87$ as sum are$ 43,44$. I hope this is of some help
