800 Followers problem!

Lets do pairing as follows:

100²-1²=101x99

99²-2²=101x97

-98²+3²= -101x95

......& so on to get 50 pairs

N=101x(99+97-95-93+91+89.......+3+1)

First alternate + & - cancels to give a result 4. This goes on till 48 terms so we get 24x4=96

We add the last +3+1 to get 100

So N=101x100=10100 & hence remainder 100 when divided by 1000

If we arrange all the numbers in a group of two as
( $100^2$ - $98^2$ )+( $99^2$ - $97^2$ )+..........( $4^2$ - $2^2$ )+( $3^2$ - $1^2$ )
=(100+98)(100-98)+(99+97)(99-97)+.............(4+2)(4-2)+(3+1)(3-1)
=2(100+98)+2(99+97)+...............2(4+2)+2(3+1)
=2(100+98+99+97+.............4+2+3+1)
=2(100+99+98+97+96+...........+5+4+3+2+1) [arranging them]
=2(5050)
=10100
Therefore remainder is 100 when divided by 1000

Consider 1st and 3rd term. $\large 100^2-98^2= 2*198, ~ so~ we ~have~~ a_n^2-a_{n+2}^2=2*(a_n+a_{n+2} )\\So~in~a~group ~of~four~we~have~ a_n^2-a_{n+2}^2+ a_{n+1}^2-a_{n+3}^2\\=2*\{a_n+a_{n+2}+ a_{n+1}+a_{n+3} \} \\\large \therefore~N= \displaystyle \sum_{k=1}^{100} 2*n_k =10100\\10100\equiv \color{#D61F06}{ 100}~ (mod~1000)$

Consider the first 4 terms

Let a = 100 , b = 99

Then , a - 2 = 98 b - 2 = 97

Therefore, the first four terms are in the form:

      a^2  + b^2  - [ ( a - 2 )^2 + ( b - 2 )^2 ]  = 4( a+b-2 )

Therefore, sum of first 4 terms is 4(100 + 99 - 2 ) = 788

Similarly, sum of next 4 terms is 756

Similarly, sum of last 4 terms is 20.

If we consider the sum of 4 consecutive terms as a single term, then we get a new AP, i.e.,

                   788, 756, .................,20

     Here a =788, n = 100/4 =25 , l = 20

( where a = first term of AP , n = no. of terms and l = last term of AP )

        Therefore,  S = n/2 ( a+ l )
                                   = 25/2 ( 788 + 20 ) 
                                 S = 10100

Therefore, when S is divided by 1000, the remainder is 100.

Hence answer is 100