This looks like a neat problem. I haven't been able to solve it, but it seems like we want to get the harmonic series in on the act somewhere. These were my thoughts:
Let $g_n=\sup_{1\leq i\leq n} f_i$. Then $g_n\nearrow F$, so by monotone convergence, $\int_\mathbb{R} F(x)dx=\lim_{n\to\infty} \int_\mathbb{R} g_n(x)dx$
Because $\sup_{x\in \mathbb{R}} f_n(x) = 1/n$ and $\int_{\mathbb{R}} f_n(x)dx = 1$, we must have that $m(\operatorname{supp}\,f_{n})\geq n$. That is, the smaller the $\sup$ is forced to be, the more spread out $f_n$ has to be if its integral will always be 1.
If we could show that for any integers $k$ and $N$, eventually the $f_n$'s spread out so much that for a sufficiently large integer $M\geq N$, $\int_{(\operatorname{supp} g_{\,N})^c} f_M\geq\frac{1}{k}$ then by an induction argument we could find a sequence of integers $d_k$ such that $\large\int_\mathbb{R} g_{d_{\,k+1}}=\int_{\operatorname{supp} g_{\,d_k}} g_{d_{\,k+1}}+\int_{(\operatorname{supp} g_{\,d_k})^c} g_{\,d_{\,k+1}}\geq\left(\sum_{j=1}^k\frac{1}{j}\right)+\frac{1}{k+1}$ so that $\int_\mathbb{R} F(x)dx=\lim_{n\to\infty} \int_\mathbb{R} g_n(x)dx=\infty.$ Perhaps this will help spur someone else to come up with the right idea, or maybe finish off my proposed argument.