Prove there are distinct $x_1,\,x_2,\cdots,\,x_n$ such that $ \sum_{i=1}^n\frac{p_i}{f'(x_i)}=\sum_{i=1}^n p_i. $ 
Suppose $f(x)$ is differentiable on $[0,\,1]$, $f(0)=0$, $f(1)=1$ and $p_1,\,p_2,\cdots,\,p_n$ are $n$ positive real numbers. Prove there are distinct $x_1,\,x_2,\cdots,\,x_n$ such that
  $$
\sum_{i=1}^n\frac{p_i}{f'(x_i)}=\sum_{i=1}^n p_i.
$$

I can only prove some special cases. 
Let $p=\sum_{i=1}^n p_i$. It suffice to prove that $\sum_{i=1}^n\frac{p_i}{pf'(x_i)}=1$. A proper choose is $f'(x_i)=\frac{np_i}{p}$. From Darboux theorem, if $f'$ is large enough, these values can attain. 

Proof. $\blacktriangleleft$ Assume $\sum p_j = 1$, otherwise replace $p_j$ by $p_j/p$ for each $j$. By continuity and the Intermediate Value Theorem, there is some $y_1 \in (0,1)$ that $f(y_1) = p_1$, then there is some $y_2 \in (y_1, 1)$ that $f(y_2) = p_1 + p_2$. Do this $n-1$ times, we obtain that 
$$
0 = y_0 < y_1 < \dots < y_{n-1} < y_n = 1, f(y_j) = \sum_{k = 1}^j p_k. 
$$
Now by the Mean Value Theorem, there is $x_j \in (y_{j-1} , y_j)$ where $f(y_j) - f(y_{j-1}) = (y_j - y_{j-1}) f'(x_j)$ for $j \leqslant n$, and
$$
\frac {p_j}{f'(x_j)} = \frac {f(y_j) - f(y_{j-1})} {f'(x_j)} = y_j - y_{j-1}, j \leqslant n, 
$$
so sum these up and we are done. $\blacktriangleright$
