Yes it deemed possible!

We could just ignore the even terms ( $2^5,4^5,...$ etc.)

If $n\equiv 1\pmod2$ , then $n\equiv n^5\pmod4$ .

So we can consider $1+3+5+...+97+99=50^2$ which is divisible by 4.

The sum is actually divisible by 100. Just observe that $a^5 + b^5 = (a+b) (a^4 - a^3 b + a^2 b^2 - ab^3 +b^4).$ Now, we can write $\begin{aligned} & 1^5 + 2^5 + \dotso + (99)^5 + (100)^5 \\ & = (100)^5 + (50)^5 + \sum_{i=1}^{49} [ i^5 + (100 - i)^5] \\ & = (100)^5 + (50)^5 + \sum_{i=1}^{49} 100 \times [ i^4 - i^3(100-i) + i^2 (100-i)^2 - i(100 - i)^3 + (100-i)^4] .\end{aligned}$ Clearly each of the terms is a multiple of 100.

Since 100 is divisible by 4, the remainder is $\large \color{#3D99F6} \boxed{0} .$

$\sum_{i=1}^{100}i^5=25 \times \sum_{i=1}^{4}i^5= 25 \times (1^5+3^5)= 25 \times (1+3)= 0 \ (mod 4)$