Like that will ever happen!

First let $A$ be the event that a fair coin is chosen at random. Then we are given that

$P(A)*(\frac{1}{2}) + P(\overline{A})*(1) = \frac{11}{17}$

$\Longrightarrow (\frac{n}{n + m})(\frac{1}{2}) + (\frac{m}{n + m})(1) = \frac{11}{17}$

$\Longrightarrow 17*(n + 2m) = 22*(n + m) \Longrightarrow 12m = 5n$ .

So $m = (\frac{5}{12})n$ . Substituting this into the given inequality gives us that

$2n \le 133 + (\frac{5}{12})n \Longrightarrow (\frac{19}{12})n \le 133 \Longrightarrow n \le 84$ .

But since $12$ and $5$ are coprime, the fact that $12m = 5n$ implies that $n$ must be a multiple of $12$ . Thus since $84 = 7*12$ , there are $7$ possible values for $n$ with $7$ corresponding values for $m$ , resulting in $\boxed{7}$ different values for $n + m$ .

(Note that since $n$ increases by increments of $12$ and $m$ increases by increments of $5$ we have that $n + m$ increases by increments of $17$ , going from a minimum of $17$ to a maximum of $119$ coins in the bag.)

The chance of head equals the total number of head, divided by the total number of sides of a coin, hence is $\frac{2m+n}{2(m+n)} = 1 -\frac{m}{2(m+n)}$ . Since this equals $\frac{11}{17}$ , we know that $\frac{m}{2(n+m)}=\frac{6}{17}$ , which reads $5n=12m$ after rearranging. Therefore the only possible solutions are for $n$ a multiple of $12$ .

Now we need to use the inequalit. We can multiply it by $12$ to obtain $24n\leq 12\cdot133 + 12m = 12\cdot133 + 5n$ , and thus $19n\leq 12\cdot133$ , or $n\leq84$ . Since $84=7\cdot12$ , there are exactly $\boxed{7}$ solutions where $m,n$ are positive.