We call a number $10^3 < n < 10^6$ a [i]balanced [/i]number if the sum of its last three digits is equal to the sum of its other digits. What is the sum of all balanced numbers in $\bmod {13}$?

$ 
\textbf{(A)}\ 0
\qquad\textbf{(B)}\ 5
\qquad\textbf{(C)}\ 7
\qquad\textbf{(D)}\ 11
\qquad\textbf{(E)}\ 12
$
1. We need to find the sum of all balanced numbers in the range \(10^3 < n < 10^6\) modulo 13. A balanced number is defined as a number where the sum of its last three digits is equal to the sum of its other digits.

2. Let's denote a balanced number as \(n = 1000a + b\), where \(a\) is the integer part of the number when divided by 1000, and \(b\) is the last three digits of the number. For \(n\) to be balanced, the sum of the digits of \(a\) must equal the sum of the digits of \(b\).

3. Consider the number \(n\) written as \(n = 1000a + b\). The sum of the digits of \(a\) is equal to the sum of the digits of \(b\). We need to find the sum of all such numbers modulo 13.

4. Suppose there are \(a_k\) three-digit numbers whose digits sum to \(k\). Then any such three-digit number \(x\) will appear \(a_k\) times after a comma in the list of balanced numbers, and \(a_k\) times before the comma, contributing \(a_k \cdot 1001 \cdot x\) to the sum of all balanced numbers.

5. Notice that \(1001 \equiv 0 \pmod{13}\) because:
   \[
   1001 \div 13 = 77 \quad \text{(exact division)}
   \]
   Therefore, \(1001 \equiv 0 \pmod{13}\).

6. Since \(1001 \equiv 0 \pmod{13}\), any number of the form \(1001 \cdot x\) will also be congruent to 0 modulo 13. Thus, the sum of all balanced numbers modulo 13 is 0.

Conclusion:
\[
\boxed{0}
\]