A(N) Problem!

Write $\frac{2014n^2+2014n+2014}{n}=2014n+2014+\frac{2014}{n}$ The first two terms are always natural for any natural number $n$ , so we'll focus with $\frac{2014}{n}$ . Since $2014=2 \times 19 \times 53$ has $(1+1)(1+1)(1+1)=8$ positive factors, then there are 8 possible natural values of $n$ for which $n$ divides 2014, that is, $\frac{2014}{n}$ is a natural number, and thus is $\frac{2014n^2+2014n+2014}{n}$ . The answer is $\boxed{8}$ .

The numerator factors as $2014\left( { n }^{ 2 }+n+1 \right)$ . Then the entire expression will be a natural number if $n$ divides either 2014 or ${ n }^{ 2 }+n+1$ evenly. Only when $n=1$ does $n$ divide the polynomial evenly. The divisors of 2014 form the set $\left\{ 1, 2, 19, 38, 53, 106, 1007, 2014 \right\}$ and the cardinality of this set is eight.

$(2014{ n }^{ 2 }+2014n+2014\\ =\quad 2014({ n }^{ 2 }+n+1)\\ now\quad n|2014({ n }^{ 2 }+n+1)\\ but\quad n\quad |\quad { n }^{ 2 }+n\\ so\quad ,\quad n\quad does\quad not\quad divide\quad { n }^{ 2 }+n+1\\ so\quad we\quad need\quad to\quad find\quad n\quad for\quad which\quad \\ n|2014\\ by\quad finding\quad out\quad its\quad factors\quad 2*19*53\\ we\quad get\quad total\quad divisors\quad of\quad 2014\quad =\quad (1+1)(1+1)(1+1)=8\quad$

$\frac{2014{n}^{2} + 2014n + 2014}{n} = \frac{2014({n}^{2} + n + 1)}{n}$

Therefore, it will be a natural number iff

2014 = 0 mod n, ${n}^{2}$ + n + 1 = 0 mod n

1st case -

It is only possible when n is a divisor of 2014. Hence $\tau(2014)$ = (1 + 1)(1 + 1)(1 + 1) [ 2014 = 2* 19 * 53]

= 8

2nd Case -

It is possible for n = 1 only for all real values(for more, check out Math Math's comment). But we have included this solution already because 1 also divides 2014.

Therefore, the answer is $\boxed{8}$