Find the sum of distinct residues of the number $2012^n+m^2$ on $\mod 11$ where $m$ and $n$ are positive integers.

$ \textbf{(A)}\ 55 \qquad \textbf{(B)}\ 46 \qquad \textbf{(C)}\ 43 \qquad \textbf{(D)}\ 39 \qquad \textbf{(E)}\ 37$
1. First, we need to simplify \(2012^n \mod 11\). Notice that:
   \[
   2012 \equiv -1 \mod 11
   \]
   Therefore,
   \[
   2012^n \equiv (-1)^n \mod 11
   \]
   This means \(2012^n \mod 11\) will be either \(1\) or \(-1\) depending on whether \(n\) is even or odd.

2. Next, we need to consider \(m^2 \mod 11\). The possible quadratic residues modulo 11 are:
   \[
   m^2 \equiv 0, 1, 4, 9, 5, 3 \mod 11
   \]
   These are the distinct quadratic residues modulo 11.

3. Now, we need to find the distinct residues of \(2012^n + m^2 \mod 11\). We consider both cases for \(2012^n \mod 11\):
   - When \(2012^n \equiv 1 \mod 11\):
     \[
     1 + 0 \equiv 1 \mod 11
     \]
     \[
     1 + 1 \equiv 2 \mod 11
     \]
     \[
     1 + 4 \equiv 5 \mod 11
     \]
     \[
     1 + 9 \equiv 10 \mod 11
     \]
     \[
     1 + 5 \equiv 6 \mod 11
     \]
     \[
     1 + 3 \equiv 4 \mod 11
     \]
   - When \(2012^n \equiv -1 \mod 11\):
     \[
     -1 + 0 \equiv 10 \mod 11
     \]
     \[
     -1 + 1 \equiv 0 \mod 11
     \]
     \[
     -1 + 4 \equiv 3 \mod 11
     \]
     \[
     -1 + 9 \equiv 8 \mod 11
     \]
     \[
     -1 + 5 \equiv 4 \mod 11
     \]
     \[
     -1 + 3 \equiv 2 \mod 11
     \]

4. Combining all distinct residues from both cases, we get:
   \[
   \{0, 1, 2, 3, 4, 5, 6, 8, 10\}
   \]

5. Finally, we sum these distinct residues:
   \[
   0 + 1 + 2 + 3 + 4 + 5 + 6 + 8 + 10 = 39
   \]

The final answer is \(\boxed{39}\).