It's nice to see less numbers

$N = \displaystyle \sum_{i=1}^n {i} = \dfrac {n(n+1)}{2}$

$\Rightarrow \dfrac {n(n+1)}{2} -(k+k+1) = 1224 \quad \Rightarrow n(n+1)-4k-2 = 2448\\ \Rightarrow k = \dfrac {n(n+1)-2450}{4} \quad \Rightarrow k$ increases as $n$ increases.

Now, let us check the smallest possible $n_{min} \approx \lfloor \sqrt{2\times 1224} \rfloor = \lfloor 49.47 \rfloor = 49$ .

We know that $k<n$ . Let us check what is the largest possible $n_{max}$ by assuming $k = n$ . Then:

$n = \dfrac {n(n+1)-2450}{4}\quad \Rightarrow 4n = n^2 + n - 2450 \\ \Rightarrow n^2 - 3n -2450 = 0 \quad \Rightarrow n = 51.02 \quad \Rightarrow n_{max} \approx 51$

Let us check the values of $k$ for $n = 49, 50, 51$ .

\(\begin{array} {} n = 49 & \Rightarrow k = \dfrac {49(50)-2450}{4} & = 0 < 1 \text{ unacceptable} \\ n = 50 & \Rightarrow k = \dfrac {50(51)-2450}{4} & = 25 \text{ acceptable} \\ n = 51 & \Rightarrow k = \dfrac {51(52)-2450}{4} & = 50.5 \text{ unacceptable} \end{array} \)

When $n > 51$ , $k > n$ , which is unacceptable. Therefore, the acceptable $k = 25$ and $k-20 = \boxed{5}$

I split the problem into two parts; finding a solution (which is fairly easy), and then showing that the solution is unique (which is a little trickier).

First of all we have from the problem

$\dfrac{n(n+1)}{2}-2k-1=1224$

$\implies k=\dfrac{n(n+1)-2450}{4}\dots(1)$

$\implies k-20=\dfrac{n(n+1)-2530}{4}\dots(2)$

Since $50\times 51=2550$ , (2) has an easy solution by inspection;

When $n=50$ , $\boxed{k-20=5}$

Now let us think about the uniqueness of this solution.

From (1) we can see that n cannot be less than 50, because k then becomes zero or negative.

From (1) we can also see that n cannot be 51 because that yields a non-integer value for k.

Finally suppose that $n=50+p$ with $p>1$ . Then (1) gives

$k=\dfrac{(50+p)(51+p)-2450}{4}=\dfrac{100+p^2+101p}{4}$

$\implies k>\dfrac{100+4p+100}{4}=50+p=n$

But this cannot be!

k cannot be omitted from the sum if it is greater than the last and greatest number in the sum.

In conclusion $(n,k)=(50,25)$ is the unique solution in positive integers and with $k<n$ to (1)