I reasoned it this way: divisibility by 4 has 4 congruence classes. Only two of these are such that $a^6 + a \equiv 2 \pmod 4$; the others have no remainder. In $\sum_{i=0}^{102} i^6+i$ there are 52 summands that belong to the first two classes. So $52.2 = 104 \equiv 0 \pmod 4$. Hence, the first remainder is 0.
Mutatis mutandis for the remainder of the division by 5, which according to my calculations is 3.
Am I too off the mark here?
On
Your method is fine. Indeed: $$\begin{align}i&\equiv0,1,2,3 \pmod{4}\\ i^2&\equiv0,1,0,1 \pmod{4}\\ i^6&\equiv0,1,0,1 \pmod{4}\\ i^6+i&\equiv0,2,2,0 \pmod{4}\end{align}\\ 4i-3\le 102 \Rightarrow i\le 26.25 \Rightarrow i=26\\ 4i-2\le 102 \Rightarrow i\le 26 \Rightarrow i=26\\ (26+26)\cdot 2=104 \equiv 0 \pmod{4}$$ Similarly: $$\begin{align}i&\equiv0,1,2,3,4 \pmod{5}\\ i^2&\equiv 0,1,-1,-1,1 \pmod{5}\\ i^6&\equiv 0,1,-1,-1,1 \pmod{5}\\ i^6+i&\equiv 0,2,1,2,0 \pmod{5}\end{align}\\ 5i-4\le 102 \Rightarrow i\le 21.2 \Rightarrow i=21\\ 5i-3\le 102 \Rightarrow i\le 21 \Rightarrow i=21\\ 5i-2\le 102 \Rightarrow i\le 20.8 \Rightarrow i=20\\ (21+20)\cdot 2+21\cdot 1=103 \equiv 3 \pmod{5}$$ Alternatively, for mod $4$: $$i^6+i=\underbrace{i(i+1)}_{0\pmod 2}(i^4-i^3+i^2-i+1)\Rightarrow \\ i^4-i^3+i^2-i+1=\underbrace{(i-1)(i^3+i)}_{0\pmod{2}}+1 \Rightarrow \\ \sum_{i=0}^{102} 1 \equiv 0\pmod{2}.$$
For $5$, the way I did it is as follows:
Note that $i^5\equiv i\mod 5$ (by Fermat's little theorem). So, $i^6+i\equiv i^2+i\mod 5$. So, your sum is congruent to $$\sum_{i=1}^{102} i^2+i=\frac{(102)(103)(205)}{6}+\frac{(102)(103)}{2}\equiv 0+3\equiv 3\mod 5.$$