Let $n{}$ be a natural number. The pairwise distinct nonzero integers $a_1,a_2,\ldots,a_n$ have the property that the number \[(k+a_1)(k+a_2)\cdots(k+a_n)\]is divisible by $a_1a_2\cdots a_n$ for any integer $k{}.$ Find the largest possible value of $a_n.$

[i]Proposed by F. Petrov and K. Sukhov[/i]
1. **Understanding the Problem:**
   We are given pairwise distinct nonzero integers \(a_1, a_2, \ldots, a_n\) such that the polynomial \((k + a_1)(k + a_2) \cdots (k + a_n)\) is divisible by \(a_1 a_2 \cdots a_n\) for any integer \(k\). We need to find the largest possible value of \(a_n\).

2. **Divisibility Condition:**
   The given condition implies that for any integer \(k\), the product \((k + a_1)(k + a_2) \cdots (k + a_n)\) must be divisible by \(a_1 a_2 \cdots a_n\). This can be interpreted as:
   \[
   \frac{(k + a_1)(k + a_2) \cdots (k + a_n)}{a_1 a_2 \cdots a_n} \in \mathbb{Z}
   \]
   for any integer \(k\).

3. **Polynomial Interpretation:**
   Consider the polynomial \(P(x) = (x + a_1)(x + a_2) \cdots (x + a_n)\). The condition implies that \(P(k)\) is divisible by \(a_1 a_2 \cdots a_n\) for any integer \(k\).

4. **Factorial Bound:**
   By the properties of polynomials and divisibility, we can infer that the product \(a_1 a_2 \cdots a_n\) must divide \(n!\). This is because the polynomial \(P(x)\) evaluated at \(x = -a_i\) (for each \(i\)) must yield a multiple of \(a_1 a_2 \cdots a_n\).

5. **Maximizing \(a_n\):**
   To maximize \(a_n\), we need to consider the arrangement of \(a_1, a_2, \ldots, a_n\) such that their product divides \(n!\). One effective way to achieve this is by choosing \(a_i\) values symmetrically around zero, i.e., \(\pm 1, \pm 2, \ldots, \pm \left\lfloor \frac{n}{2} \right\rfloor\).

6. **Odd and Even \(n\):**
   - For odd \(n\), the largest \(a_n\) can be \(\frac{n!}{\left(\left\lfloor \frac{n}{2} \right\rfloor!\right)^2}\).
   - For even \(n\), the largest \(a_n\) can be \(\frac{n! \cdot \left\lfloor \frac{n}{2} \right\rfloor}{\left(\left\lfloor \frac{n}{2} \right\rfloor!\right)^2}\).

7. **Verification:**
   To verify that this construction works, we use the properties of the floor function and \(v_p\) (p-adic valuation) manipulations to show that the product of the chosen \(a_i\) values indeed divides \(n!\).

The final answer is \(\boxed{\frac{n!}{\left(\left\lfloor \frac{n}{2} \right\rfloor!\right)^2}}\) for odd \(n\) and \(\boxed{\frac{n! \cdot \left\lfloor \frac{n}{2} \right\rfloor}{\left(\left\lfloor \frac{n}{2} \right\rfloor!\right)^2}}\) for even \(n\).