To solve this problem, we will analyze the given condition involving permutations and perfect squares to determine the smallest $n$ such that $P(n)$, the number of permutations of $[n] = \{1, 2, \ldots, n\}$ where $ka_k$ is a perfect square for all $1 \leq k \leq n$, is a multiple of 2010.

### Step-by-Step Analysis

1. Understanding the Problem:

Given a permutation $(a_1, a_2, \ldots, a_n)$ of $[n]$, we need each product $ka_k$ to be a perfect square. This implies $ka_k = m_k^2$ for some integer $m_k$.

Therefore, $a_k = \frac{m_k^2}{k}$ must be an integer. Hence, $k$ must divide $m_k^2$. Since $m_k^2 = ka_k$, this implies that $a_k$ must also divide $k$.

2. Condition Analysis:

The divisibility condition reduces to:
$$a_k = \frac{m_k^2}{k}$$
implying $k \mid m_k^2$. This is equivalent to saying that $k$ must be a perfect square itself, because for $a_k$ to be a positive integer permutation of 1 to $n$, $m_k = \sqrt{k}$ is the simplest choice, allowing $k$ to divide $m_k^2 = k$.

3. **Valid $n$ for a Permutation:**

Next, for which values of $n$ can we construct permutations meeting the conditions? Each $k$ must be a perfect square, so $1, 4, 9, 16, \ldots$ need to be the indices selected for permutation.

4. Counting the Permutations:

First, we need to determine how many perfect squares exist within the set $[n]$. Let this count be denoted as $f(n)$, the floor of the square root of $n$:
$$f(n) = \lfloor \sqrt{n} \rfloor$$

For $P(n)$ to be non-zero, each $k$ must be a perfect square up to $n$. The constraint on $f(n)$ determining permutations is that it needs to reach a number such that the product of the factorials of the counts of solution possibilities is a multiple of 2010.

5. **Finding the Minimum $n$:**

We need:
$$P(n) = f(n)! \equiv 0 \pmod{2010}$$

Prime Factorization of 2010:
$$2010 = 2 \times 3 \times 5 \times 67$$
The smallest factorial $f(n)!$ has at least these factors.

6. **Calculating $n$:**

Approximate $f(n)!$ for increasing $n$ (especially its factorial incremental):

- The smallest $f(n)$ where $f(n)!$ is divisible by 67 is when $f(n) \approx 67$ because the smallest factorial value divisible by 67 is $67!$.

Finding where the number of perfect squares, $f(n)$, equals 67 should give us the smallest $n$:
$$n = (67)^2 = 4489$$

Thus, the smallest $n$ such that $P(n)$ is a multiple of 2010 is:
$$\boxed{4489}$$