Oh yes - i meant positive integers...I changed it from the original "natural numbers" - but accidentally wrote nonnegative...

Call the minimum possible value of the expression $d$. Obviously, $d \leq{14}$, because $14 = |2011^{1}-45^{2}|$. Note also that $d$ is even, as the expression itself is even. Also, $d$ is not divisible by $3,5$. Hence, the only possible values for $d$ are $2,4,8,14$. Note also that $2011^{m}$ is congruent to $1 mod 5$, $45^{n}=0(mod 5)$ for any $m$, so $d=1(mod 3)$. It means that either $d=1(mod 5)$ or $d=4(mod 5)$. It means that values $2,8$ are eliminated from the possible numbers' list of $d$. Now, $d=4$ or $d=14$. Assume $d=4$. Consider 2 cases: 1) $2011^{m}-45^{n}=4$ For this case, obviously $45^{n}=1(mod 4)$ and therefore, $2011^{m}=(-1)^m (mod 4)$. So, in order to have $(-1)^m=1(mod 4)$, we must take $m=2a$ for some positive integer $a$. $ (2011^{a}-2) (2011^{a}+2) = 3^{2n} \cdot {5^{n}}$. From the last equality, it's clear that $2011^{a}-2$ and $2011^{a}+2$ differ by $4$, so one of them should be divisible by $3^{2n}$ and the other one by $5^{n}$, so $9^n-2^{n}=4$ which gives $n=1$ but then $a$ doesn't exist. 2) $45^{n}-2011^{m}=4$, or $45^{n}-4=2011^{m}$ $45^{n}-4=2(mod 3)$, but $2011^{m}=1(mod 3)$. Hence, both cases are impossible and the minimal possible value is \[14\].

Easy. Just use mod 5 and 3 in two cases to get that the minimum is 14 for m=1, n=2.