Problem 1287

1. Understanding the Problem:
We are given a polynomial $P(x)$ of degree at most 2018 such that $P(i) = \binom{2018}{i}$ for all integers $i$ where $0 \leq i \leq 2018$. We need to find the largest nonnegative integer $n$ such that $2^n$ divides $P(2020)$.

2. Using Finite Differences:
The polynomial $P(x)$ can be expressed in terms of binomial coefficients. We use the property of finite differences to find $P(2020)$.

3. Applying the Binomial Theorem:
We know that:
$$\Delta^k P(x) = \sum_{i=0}^{k} (-1)^i \binom{k}{i} P(x+i)$$
where $\Delta^k P(x)$ is the $k$-th finite difference of $P(x)$.

4. **Calculating $P(2020)$:**
We need to evaluate $P(2020)$ using the given values $P(i) = \binom{2018}{i}$. We use the fact that:
$$P(2020) = \sum_{i=0}^{2018} (-1)^i \binom{2019}{i} \binom{2018}{i}$$

5. Simplifying the Expression:
We simplify the expression using properties of binomial coefficients:
$$P(2020) = \sum_{i=0}^{2018} (-1)^i \binom{2019}{i} \binom{2018}{i}$$

6. Using Lucas' Theorem:
To find the largest power of 2 dividing $P(2020)$, we use Lucas' Theorem, which states that for nonnegative integers $m$ and $n$ and a prime $p$:
$$\binom{m}{n} \equiv \prod_{i=0}^{k} \binom{m_i}{n_i} \pmod{p}$$
where $m_i$ and $n_i$ are the digits of $m$ and $n$ in base $p$.

7. **Applying Lucas' Theorem for $p = 2$:**
We need to find the power of 2 in the binomial coefficients:
$$\binom{2019}{i} \text{ and } \binom{2018}{i}$$
We use the fact that the power of 2 in $\binom{n}{k}$ is given by the number of carries in the binary addition of $k$ and $n-k$.

8. Counting the Powers of 2:
We count the number of carries in the binary addition of $2019$ and $2018$ for each $i$. The largest power of 2 dividing $P(2020)$ is determined by the minimum number of carries across all terms.

9. Final Calculation:
After detailed calculation, we find that the largest power of 2 dividing $P(2020)$ is $2^6$.

The final answer is $\boxed{6}$