If $\ \sum_{k=1}^n m(E_n) > n-1,$ then prove that $\bigcap_{k=1}^n E_k$ has positive measure.

Let $A=\bigcap_{k=1}^nE_k$

Assume that $m(A)=m(\bigcap_{k=1}^nE_k) =0 $.

Then $1=m([0,1]$ \ $A) \leq \sum_{k=1}^nm([0,1]$ \ $E_k) =n -\sum_{k=1}^nm(E_k) $

From this we see that $$\sum_{k=1}^nm(E_k) \leq n-1$$

which is a contradiction.

The following is a direct proof, credited to Marios.

Observe that $$m\left([0,1] \setminus \bigcap_{k=1}^n E_k \right) \leq \sum_{k=1}^nm([0,1] \setminus E_k) = n - \sum_{k=1}^n m(E_k) < n - (n-1) = 1.$$ Therefore, $$m\left( \bigcap_{k=1}^n E_k \right) > 0 .$$