Beginning

For even $n$ , we know that $n^5$ is divisible by $2^5=32$ . For odd $n$ , observe that $n^5\equiv{n}\pmod{4}$ . Thus $\sum_{k=1}^{50}(2k-1)^5\equiv\sum_{k=1}^{50}(2k-1)=50^2\equiv{0}\pmod{4}$ . Thus the remainder is $\boxed{0.}$

Bonus question: What is the remainder when the sum is divided by 16?

I think most people have done it by fluke....and got their ans. Correct... I mean is it really of lvl 2 type

Grouping the value in terms of four gives $x^5+(x+1)^5+(x+2)^5+(x+3)^5 = 4(x^5+35x^3+90x^2+69)+10x(3x^2+49)$ The hardest part is find the reminder by dividing $10x(3x^3+49)$ with 4. For $x$ in even number, it is obvious that $10x$ is dividable by 4. $x$ in odd number, when cube, also remains odd and when add with odd (49), will result to the even number, making the whole value dividable by 4 as well, thus, thus, the reminder will be 0. Repeat the process for $x=1,5,9,13,\dots,97$ also yields the same result, thus $\sum_{x=1}^{100}(x^5)$ is dividable by 4 and the remainder is 0.

Group data by group of four elements from whole set of number: {1,2,3,4}, {5,6,7,8},...,{97,98,99,100}. It is easy to see each group has remainder 0 when dividing by 4. So: $ans=\boxed{0}$