To find the greatest common divisor (GCD) of the integers $P(1), P(2), P(3), \ldots, P(2016)$, where $P(n) = n(n+1)(2n+1)(3n+1)\cdots(16n+1)$, we will first consider each part of the product and determine if there is a consistent factor across all $P(n)$.

### Step 1: Analyze the Form of $P(n)$

The expression for $P(n)$ involves the product:

$$P(n) = n(n+1)(2n+1)(3n+1)\cdots(16n+1)$$

To find a common factor, we need to explore each factor modulo small primes.

### Step 2: Explore Modulo Small Primes

We'll compute $P(n)$ modulo small primes to find a potential common divisor. The primary candidates are small primes.

#### Consider modulo 2:

- $n \equiv 0 \pmod{2}$ or $n+1 \equiv 0 \pmod{2}$, so for all $n$, $P(n) \equiv 0 \pmod{2}$.

#### Consider modulo 3:

- For any $n$: $n \equiv 0, 1, 2 \pmod{3}$, one of the factors $n, (n+1), (2n+1), \ldots, (16n+1)$ will be divisible by 3. Thus, $P(n) \equiv 0 \pmod{3}$.

#### Consider modulo 5:

- For any $n$, examine possible values of each factor modulo 5. One of $n, (n+1), (2n+1), (3n+1), \ldots, (16n+1)$ will be divisible by 5 over five consecutive values, thus $P(n) \equiv 0 \pmod{5}$.

By checking analogous conditions for each prime factor:

- Modulo 7: Similarly, there exists at least one factor of 7.
- Modulo 11: Similarly, there exists at least one factor of 11.
- Modulo 13: Similarly, there exists at least one factor of 13.
- Modulo 17: Similarly, there exists at least one factor of 17.

### Step 3: Conclude the GCD

The GCD of $P(1), P(2), P(3), \ldots, P(2016)$ is the product of these common factors across $n$:

$$\boxed{510510}$$

This number, 510510, is $2 \times 3 \times 5 \times 7 \times 11 \times 13 \times 17$, the product of the primes up to and including 17, ensuring it divides each $P(n)$.