Calculate the sum of the digits of the first 100 numbers of that sequence which are divisible by 202.​

Note that $n=\sum_{k=0}^r c_k\cdot 100^k$ is divisible by $101$ if and only if $$n=\sum_{k=0}^r c_k\cdot 100^k\equiv \sum_{k=0}^rc_k(-1)^k\equiv 0 \pmod{101}.$$ In other words, given a number $n$ take the digits in pairs from the right and alternately add and subtract them. The number $n$ is divisible by $101$ if and only if the result is zero modulo $101$ (it reminds the divisibility rule for $11$)

Therefore in your sequence, a number is divisible by $202=2\cdot 101$ iff the number of its digits is $0\pmod 4$ or $3\pmod 4$: $$202,2020,2020202,20202020,20202020202,202020202020\dots$$

So the sum of the digits of the first 100 numbers of that sequence which are divisible by $202$ is $$4\cdot(1+1+2+2+3+3+\dots+50+50)$$ What is the final answer?