Let number of removed cards are $$k$$ and $$k+1$$.

Given, the sum of numbers on the cards after removing $$k$$ and $$k+1$$ is 1224.

$$\begin{aligned} & \therefore(1+2+3+\ldots .+n)-(k+(k+1))=1224 \\ & \Rightarrow \quad \frac{n(n+1)}{2}-2 k-1=1224 \\ & \Rightarrow \quad \frac{n(n+1)}{2}-2 k=1225 \\ & \Rightarrow \quad n^2+n-4 k=2450 \\ & \Rightarrow \quad n^2+n-2450=4 k \\ & \Rightarrow(n+50)(n-49)=4 k \\ & \therefore \quad n>49 \\ & \text { Let } n=50 \\ & \Rightarrow \quad 100 \times 1=4 k \\ \end{aligned}$$

$$\begin{aligned} \Rightarrow \quad k =25 \\ \Rightarrow \quad k-20 =5 \end{aligned}$$

Hints :

(i) Recall $$1+2+3+\ldots+n=\frac{n(n+1)}{2}$$

(ii) If $$k$$ is the smallest number on two consecutive numbers, then the second number is $$k+1$$.