4. Let $f(x)=\frac{1}{x^{3}+3 x^{2}+2 x}$. Determine the smallest positive integer $n$ such that
$$
f(1)+f(2)+f(3)+\cdots+f(n)>\frac{503}{2014}
$$
Solution 1: Note that $x^{3}+3 x^{2}+2 x=x(x+1)(x+2)$. We first write $1 /\left(x^{3}+3 x^{2}+2 x\right)$ as
$$
\frac{a}{x}+\frac{b}{x+1}+\frac{c}{x+2}
$$
where $a, b, c$ are real numbers.
This expression simplifies to
$$
\frac{a(x+1)(x+2)+b x(x+2)+c x(x+1)}{x(x+1)(x+2)}=\frac{(a+b+c) x^{2}+(3 a+2 b+c) x+2 a}{x(x+1)(x+2)} .
$$

Hence, $a+b+c=0,3 a+2 b+c=0$ and $2 a=1$.
From the last equation, we get $a=1 / 2$. The first two equations simplifies to $b+c=-1 / 2$ and $2 b+c=-3 / 2$. Subtracting the former from the latter yields $b=-1$. Hence, $c=1 / 2$. Therefore,
$$
f(x)=\frac{1}{2}\left(\frac{1}{x}-\frac{2}{x+1}+\frac{1}{x+2}\right) .
$$

We now focus on the sum $f(1)+f(2)+f(3)+\ldots+f(n)$. This is equal to
$$
\frac{1}{2}\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n}\right)-\left(\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n+1}\right)+\frac{1}{2}\left(\frac{1}{3}+\frac{1}{4}+\ldots+\frac{1}{n+2}\right) .
$$

All terms except $1,1 / 2,1 /(n+1), 1 /(n+2)$ cancel out completely. Hence, the result simplifies to
$$
\frac{1}{2}-\frac{1}{4}-\frac{1}{2} \cdot \frac{1}{(n+1)}+\frac{1}{2} \cdot \frac{1}{n+2}=\frac{1}{4}-\frac{1}{2(n+1)(n+2)} .
$$

We can now solve for the desired answer, i.e. find the smallest positive integer $n$ such that
$$
\frac{1}{4}-\frac{1}{2(n+1)(n+2)}>\frac{503}{2014} \text {. }
$$

This simplifies to
$$
\frac{1}{2(n+1)(n+2)}2014$.
By trial and error and estimation, we see that $44 \cdot 45=1980$ and $45 \cdot 46=2070$. Hence, $n=44$ is the desired answer.
Solution 2: We obtain $(n+1)(n+2)>2014$ as in Solution 1. This simplifies to $n^{2}+3 n-$ $2012>0$. By the quadratic formula, we obtain
$$
n>\frac{-3+\sqrt{3^{2}+4 \cdot 2012}}{2}=\frac{-3+\sqrt{8057}}{2} \text {. }
$$

The largest positive integer less than $\sqrt{8057}$ is 89 . Hence, $n>(-3+89) / 2=43$. Therefore, $n=44$ is the smallest positive integer satisfying the given eqation.