Let $(a_n)_{n\geq 1}$ be a sequence such that $a_1=1$ and $3a_{n+1}-3a_n=1$ for all $n\geq 1$.  Find $a_{2002}$.

$\textbf{(A) }666\qquad\textbf{(B) }667\qquad\textbf{(C) }668\qquad\textbf{(D) }669\qquad\textbf{(E) }670$
1. We start with the given recurrence relation:
   \[
   3a_{n+1} - 3a_n = 1
   \]
   Simplifying, we get:
   \[
   a_{n+1} - a_n = \frac{1}{3}
   \]

2. To find a general formula for \(a_n\), we can express \(a_{n+1}\) in terms of \(a_n\):
   \[
   a_{n+1} = a_n + \frac{1}{3}
   \]

3. We know the initial condition \(a_1 = 1\). Let's find the first few terms to identify a pattern:
   \[
   a_2 = a_1 + \frac{1}{3} = 1 + \frac{1}{3} = \frac{4}{3}
   \]
   \[
   a_3 = a_2 + \frac{1}{3} = \frac{4}{3} + \frac{1}{3} = \frac{5}{3}
   \]
   \[
   a_4 = a_3 + \frac{1}{3} = \frac{5}{3} + \frac{1}{3} = 2
   \]

4. Observing the pattern, we hypothesize that:
   \[
   a_n = 1 + \frac{n-1}{3}
   \]
   Simplifying, we get:
   \[
   a_n = \frac{3 + n - 1}{3} = \frac{n + 2}{3}
   \]

5. To verify, we use induction:
   - **Base case**: For \(n = 1\),
     \[
     a_1 = \frac{1 + 2}{3} = 1
     \]
     which is true.
   - **Inductive step**: Assume \(a_k = \frac{k + 2}{3}\) holds for some \(k \geq 1\). Then,
     \[
     a_{k+1} = a_k + \frac{1}{3} = \frac{k + 2}{3} + \frac{1}{3} = \frac{k + 3}{3}
     \]
     which matches the form \(\frac{(k+1) + 2}{3}\).

6. By induction, the formula \(a_n = \frac{n + 2}{3}\) is valid for all \(n \geq 1\).

7. To find \(a_{2002}\):
   \[
   a_{2002} = \frac{2002 + 2}{3} = \frac{2004}{3} = 668
   \]

The final answer is \(\boxed{668}\)