Let $N$ be the number of (positive) divisors of $2010^{2010}$ ending in the digit $2$. What is the remainder when $N$ is divided by 2010?
1. **Factorization and Units Digit Analysis**:
   We start by noting that \(2010^{2010} = 2^{2010} \cdot 3^{2010} \cdot 5^{2010} \cdot 67^{2010}\). Since we are interested in the units digit of the divisors, we can ignore the factor \(5^{2010}\) because any number ending in 0 or 5 will not end in 2. Thus, we focus on the product \(2^a \cdot 3^b \cdot 67^c\) where \(a, b, c\) are nonnegative integers.

2. **Units Digit Cycles**:
   - The units digit of \(2^a\) cycles every 4: \(2, 4, 8, 6\).
   - The units digit of \(3^b\) cycles every 4: \(3, 9, 7, 1\).
   - The units digit of \(67^c\) cycles every 4: \(7, 9, 3, 1\).

3. **Case Analysis**:
   We need to find combinations of \(a, b, c\) such that the product \(2^a \cdot 3^b \cdot 67^c\) ends in 2. We analyze each possible units digit of \(2^a\):

   - **Case 1**: Units digit of \(2^a\) is 2.
     - Possible pairs for \(3^b\) and \(67^c\) are: (3, 7), (9, 9), (7, 3), (1, 1).
     - Number of divisors: \(503 \cdot (503^2 + 503^2 + 502^2 + 503^2) = 503 \cdot (3 \cdot 503^2 + 502^2)\).

   - **Case 2**: Units digit of \(2^a\) is 4.
     - Possible pairs for \(3^b\) and \(67^c\) are: (3, 1), (9, 7), (7, 9), (1, 3).
     - Number of divisors: \(503 \cdot (503^2 + 503^2 + 502 \cdot 503 + 503 \cdot 502) = 503 \cdot 502 \cdot 2010\).

   - **Case 3**: Units digit of \(2^a\) is 8.
     - Possible pairs for \(3^b\) and \(67^c\) are: (3, 3), (9, 1), (7, 7), (1, 9).
     - Number of divisors: \(502 \cdot (503 \cdot 502 + 503^2 + 502 \cdot 503 + 503^2) = 502 \cdot 503 \cdot 2010\).

   - **Case 4**: Units digit of \(2^a\) is 6.
     - Possible pairs for \(3^b\) and \(67^c\) are: (3, 9), (9, 3), (7, 1), (1, 7).
     - Number of divisors: \(502 \cdot (503^2 + 503 \cdot 502 + 502 \cdot 503 + 503^2) = 502 \cdot 503 \cdot 2010\).

4. **Modulo Calculation**:
   - The number of divisors in cases 2, 3, and 4 are divisible by 2010, so they do not contribute to the remainder.
   - We focus on the result from Case 1: \(503 \cdot (3 \cdot 503^2 + 502^2)\).

5. **Simplifying the Expression**:
   \[
   3 \cdot 503^2 + 502^2 = (503^2 + 502^2) + 2 \cdot 503^2
   \]
   \[
   = (503 + 502)^2 - 2 \cdot 503 \cdot 502 + 2 \cdot 503^2
   \]
   \[
   = 1005^2 + 2 \cdot 503
   \]
   \[
   = 1004 \cdot 1005 + 1005 + 1006 \equiv 1005 + 1006 \pmod{2010}
   \]
   \[
   \equiv 1 \pmod{2010}
   \]
   Therefore, the number of divisors is \(503 \cdot 1 = 503\).

The final answer is \(\boxed{503}\).