Since
$$\begin{aligned} n^{1003} + n^{1002} + n^{1001} + 1 &= n^{1001}(n^2 + 1) + n^{1002} + 1 \\ &= n^{1001}(n^2 + 1) + (n^2 + 1)(n^{1000} - n^{998} + n^{996} - \dots - n^2 + 1) \end{aligned}$$
hence, the numerator is divisible by $n^2 + 1$. We can see that the numerator is also divisible by $n+1$, as follows from the equation:
$$\begin{aligned} n^{1003} + n^{1002} + n^{1001} + 1 &= n^{1002}(n+1) + n^{1001} + 1 \\ &= n^{1002}(n+1) + (n+1)(n^{1000} - n^{998} + n^{996} - \dots - n + 1). \end{aligned}$$

$$\frac{n^{1003} + n^{1002} + n^{1001} + 1}{n+1} = \frac{(n+1)(n^2+1)P(n)}{n+1} = (n^2+1)P(n).$$
It is easy to see that $P(n) > 1$. The statement has been proven.