Numbers $x,y,z$ are positive integers and satisfy the equation $x+y+z=2013$. (E)
a) Find the number of the triplets $(x,y,z)$ that are solutions of the equation (E).
b) Find the number of the solutions of the equation (E) for which $x=y$.
c) Find the solution $(x,y,z)$  of the equation (E) for which the product $xyz$ becomes maximum.
### Part (a)
1. We need to find the number of positive integer solutions to the equation \(x + y + z = 2013\).
2. We can use the stars and bars method to find the number of solutions to the equation \(x + y + z = 2013\) where \(x, y, z\) are positive integers.
3. The stars and bars theorem states that the number of solutions to the equation \(x_1 + x_2 + \cdots + x_k = n\) in positive integers is given by \(\binom{n-1}{k-1}\).
4. Here, \(n = 2013\) and \(k = 3\), so the number of solutions is:
   \[
   \binom{2013-1}{3-1} = \binom{2012}{2}
   \]
5. Calculating \(\binom{2012}{2}\):
   \[
   \binom{2012}{2} = \frac{2012 \cdot 2011}{2} = 2012 \cdot 1006 = 2,023,066
   \]

### Part (b)
1. We need to find the number of solutions to the equation \(x + y + z = 2013\) where \(x = y\).
2. Substituting \(x = y\) into the equation, we get:
   \[
   2x + z = 2013
   \]
3. Rearranging, we find:
   \[
   z = 2013 - 2x
   \]
4. For \(z\) to be a positive integer, \(2013 - 2x > 0\), which implies:
   \[
   2x < 2013 \implies x < 1006.5
   \]
5. Since \(x\) must be a positive integer, \(x\) can take values from 1 to 1006.
6. Therefore, there are 1006 possible values for \(x\), and for each \(x\), there is a corresponding \(z\) that is odd.
7. Thus, the number of solutions is:
   \[
   \boxed{1006}
   \]

### Part (c)
1. We need to find the solution \((x, y, z)\) of the equation \(x + y + z = 2013\) for which the product \(xyz\) is maximized.
2. By the Arithmetic Mean-Geometric Mean (AM-GM) inequality, we have:
   \[
   \frac{x + y + z}{3} \geq \sqrt[3]{xyz}
   \]
3. Equality holds when \(x = y = z\).
4. Setting \(x = y = z\), we get:
   \[
   3x = 2013 \implies x = \frac{2013}{3} = 671
   \]
5. Therefore, the solution that maximizes \(xyz\) is:
   \[
   \boxed{(671, 671, 671)}
   \]