Prove that there exist $x_1,x_2,...,x_{100}\in(1918,2018)$ such that $\sum_{k=1}^{100}f'(x_k)=100$

Question

Let $f:[1918,2018]\rightarrow\mathbb{R}$, continuous over $[1918,2018]$, differentiable on$(1918,2018)$ and such that $f(1918)=1918,f(2018)=2018$. $a)$Prove that there exist $x_1,x_2,...,x_{100}\in(1918,2018)$ such that $\sum_{k=1}^{100}f'(x_k)=100$ and if $b)$ $f$ is strictly increasing, show that there exist $y_1,y_2,...,y_{100}\in(1918,2018)$ such that $\sum_{k=1}{100}\frac{1}{y_k}=100$. I have chosen a function $h(x)=f(x)-1919$ such that $h(1918)=-1$, $h(2018)=99$ and from Bolzano's lemma there is $x_1$ in $(1918,2018)$ such that $h(x_1)=0$, so $f(x_1)=1919$. Also, from Lagrange's theorem there is at least one $x_1$, different from the previous one, between $1918$ and $2018$ such that $f(x_1)=\frac{f(2018)-f(1918)}{100}=1$. How can I extend this to show that I can find similar points in which the function is equal to $1$ such that I obtain $100$ by summing them?