$1,2,3,4, \ldots , n$ is the set of first $n$ natural number.

Now we take out any two consecutive number (say $k$ and $k+1$ ) from the list such that the average of the remaining numbers is 26.25.

Find the value of $n + k .$

**
Clarification
**
: The value of
$k$
is the smallest of the two consecutive integers.

Average of $1$ and $2$ is $\frac {1+2}{2} = 1.5 .$

The answer is 57.

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From the question it can be written as:

$\dfrac{n(n+1)}{2}-(2k+1)=\dfrac{105}{4} (n-2)$

After simplication :

$2 n^2-103n+206=8 k$

So $8 | 2 n^2-103n+206$

$8| 2 n^2- 3 n +6$

If $n$ is odd there is no chance to be divisible by 8 because whole expression will be odd.

If $n$ is even then $2 n^2$ will be divisible by 8.The remaining part $3 n-6$ should be divisible by 8.

So, $8| 3 n-6$

Now put $n=8m+r$ where $0 \leq r \leq 7$

It can be seen only $n=8m+2$ satisfy.

Then putting $n=8m+2$ in the equation we get :

$16m^2-95m+1=k$

As $k \gt 0$ then $16m^2-95m \gt 0$ .It implies $m \geq 6$

Putting $m=6$ in the equation we get $k=7$ and $n=50$

So answer is $\boxed {57}$