This was a relatively simple puzzle and the number of people who sent correct answers was overwhelming. I will try to capture as many as I can – Ishir Gupta (only one not to use algebra), Pratik Poddar, Suman Saraf, Kunal Goyal, Girish Tutakne, Karan Sharma, Abhinav Jain, Abhinav Kumar Gupta, Siddharth Goel, Chandu Karan, Manphul Budaniya, Ravi Kabra, Anupam Sarawagi and Garima Rai. Well done all – and thank you for your continued support and interest in this puzzle blog that keeps me going!

I like the clarity and detail in Abhinav Jain’s solution, though most people sent similar answers, but I am reproducing it (another answer later):

Let the no. be x
x>2000 (there is no such no. Between 1978-2000)
I am assuming x to be a 4 digit no.
1000a + 100b+ 10c + d = x
And
(10a + b) + (10c + d) = 10b + c
10a + d = 9(b-c)
Therefor 10a + d is a multiple of 9.
Where :
a isn’t equal to 0.
1<a<10
0<= b,c,d < 10
b > c
The possible values of 10a + d are :
1. 27    a=2  d=7
2. 36    a=3.  d=6
3. 45.  a=4.  d=5
4. 54.  a=5.  d=4
5. 63.  a=6.  d=3
6. 72.  a=7.  d=2
7. 81.  a=8.  d=1

Since we have to find the immediate next no.
a=2    d=7
Therefor
b-c=3
We have to find the least value of b and c for which the equation is true
b=3
c=0

Therefor the no. Is 2307
And the next no. Will be 2417, 2527,2637………… The last 4- digit no. Will be
8901.

I also like the solution sent by Ishir, even though it is missing one step (of why there is a carry over), but I really like the thinking and it is done without algebra: