The smaller number removed = k

The next number removed = k + 1

 $\therefore$        1 + 2 + 3 + .......... + n = (k) + (k + 1) + 1224

 $\frac{\mathrm{n}(\mathrm{n}+1)}{2}$  = 2k + 1225

 $\Rightarrow$       n² + n = 4k + 2450

 $\Rightarrow$      n² + n – 2450 = 4k

 $\Rightarrow$       (n + 50)(n – 49) = 4k

Here   1 < k < n and either of (n + 50) or (n – 49)

must be a multiple of '4' as

because if n is odd then (n – 49) is even

and if n is even then (n + 50) is even

so, for n = 50, k = 25

but for n = 53 ; k = 103    $\Rightarrow$         k > n  (not allowed)

rest values of n are not allowed

 $\therefore$        k = 25 and k – 20 = 5