2. Let positive integers $m, n$ satisfy
$$
m(n-m)=-11 n+8 \text {. }
$$

Then the sum of all possible values of $m-n$ is $\qquad$
2. 18 .

From the problem, we have
$$
n=\frac{m^{2}+8}{m+11}=m-11+\frac{129}{m+11} \in \mathbf{Z}_{+} \text {. }
$$

Then $(m+11) \mid 129$
$$
\Rightarrow m+11=1,3,43,129 \text {. }
$$

Also, $m \in \mathbf{Z}_{+}$, checking we find that when $m=32,118$, the corresponding $n$ is a positive integer.
$$
\text { Hence }(m, n)=(32,24) \text { or }(118,108) \text {. }
$$

Therefore, the sum of the possible values is
$$
(32-24)+(118-108)=18 \text {. }
$$