Let $F_m$ be the $m$'th Fibonacci number, defined by $F_1=F_2=1$ and $F_m = F_{m-1}+F_{m-2}$ for all $m \geq 3$. Let $p(x)$ be the polynomial of degree 1008 such that $p(2n+1)=F_{2n+1}$ for $n=0,1,2,\ldots,1008$. Find integers $j$ and $k$ such that $p(2019) = F_j - F_k$.

1. **Define the polynomial and the transformation:**
   Let \( N = 1009 \). Define \( q(x) = p(2x+1) \) so that \( q(k) = F_{2k+1} \) for \( k = 0, 1, 2, \ldots, N-1 \). We need to find \( q(N) \).

2. **Degree of the polynomial:**
   Since \( \deg q = N-1 \), we have that \( \Delta^N q \equiv 0 \). Therefore,
   \[
   q(N) = q(N) - \Delta^N q(0) = \sum_{k=0}^{N-1} (-1)^k \binom{N}{k} q(k) = \sum_{k=0}^{N-1} (-1)^k \binom{N}{k} F_{2k+1}.
   \]

3. **Define the sum \( S_n \):**
   Let \( S_n \) be the sum on the right with \( N \) replaced by \( n \), and \( S_{n,a} \) be the sum with \( F_{2k+1} \) replaced by \( F_{2k+a} \). Then,
   \[
   S_{n,a} = \sum_{k=0}^{n-1} (-1)^k \binom{n}{k} F_{2k+a}.
   \]

4. **Simplify \( S_{n,a} \):**
   Using the binomial coefficient identity \( \binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k} \), we get:
   \[
   S_{n,a} = \sum_{k=0}^{n-1} (-1)^k \left[ \binom{n-1}{k-1} + \binom{n-1}{k} \right] F_{2k+a}.
   \]
   This can be split into two sums:
   \[
   S_{n,a} = -\sum_{k=1}^{n-1} (-1)^{k-1} \binom{n-1}{k-1} F_{2k+a} + \sum_{k=0}^{n-1} (-1)^k \binom{n-1}{k} F_{2k+a}.
   \]
   Shifting the index in the first sum:
   \[
   S_{n,a} = -\sum_{k=0}^{n-2} (-1)^k \binom{n-1}{k} F_{2k+2+a} + \sum_{k=0}^{n-1} (-1)^k \binom{n-1}{k} F_{2k+a}.
   \]
   Combining the sums:
   \[
   S_{n,a} = (-1)^{n-1} F_{2n-2+a} - \sum_{k=0}^{n-2} (-1)^k \binom{n-1}{k} [F_{2k+2+a} - F_{2k+a}].
   \]
   Using the Fibonacci recurrence relation \( F_{2k+2+a} - F_{2k+a} = F_{2k+1+a} \):
   \[
   S_{n,a} = (-1)^{n-1} F_{2n-2+a} - \sum_{k=0}^{n-2} (-1)^k \binom{n-1}{k} F_{2k+1+a}.
   \]

5. **Special case for \( S_n \):**
   For \( n \geq 2 \):
   \[
   S_n = S_{n,1} = (-1)^{n-1} F_{2n-1} - \sum_{k=0}^{n-2} (-1)^k \binom{n-1}{k} F_{2k+2}.
   \]
   Using the Fibonacci recurrence relation again:
   \[
   S_n = (-1)^{n-1} F_{2n-1} - \sum_{k=0}^{n-2} (-1)^k \binom{n-1}{k} [F_{2k+1} + F_{2k}].
   \]
   Splitting the sum:
   \[
   S_n = (-1)^{n-1} F_{2n-1} - \sum_{k=0}^{n-2} (-1)^k \binom{n-1}{k} F_{2k+1} - \sum_{k=0}^{n-2} (-1)^k \binom{n-1}{k} F_{2k}.
   \]
   Recognizing the sums:
   \[
   S_n = (-1)^{n-1} F_{2n-1} - S_{n-1} - S_{n-1,0}.
   \]
   Using the previous result for \( S_{n-1,0} \):
   \[
   S_{n-1,0} = (-1)^{n-2} F_{2n-4} - \sum_{k=0}^{n-3} (-1)^k \binom{n-2}{k} F_{2k+1}.
   \]
   Combining:
   \[
   S_n = (-1)^{n-1} F_{2n-1} - S_{n-1} - (-1)^{n-2} F_{2n-4} + \sum_{k=0}^{n-3} (-1)^k \binom{n-2}{k} F_{2k+1}.
   \]
   Simplifying:
   \[
   S_n = -S_{n-1} + S_{n-2} + (-1)^{n-1} [F_{2n-1} + F_{2n-4}].
   \]

6. **Induction base cases:**
   With \( S_0 = 0 \) and \( S_1 = 1 \), an easy induction (using the Fibonacci definition) gives:
   \[
   S_n = (-1)^{n-1} (F_{2n+1} - F_{n+1}).
   \]

7. **Final calculation:**
   In particular, \( p(2019) = q(N) = S_N = F_{2019} - F_{1010} \).

Conclusion:
\[
(j, k) = \boxed{(2019, 1010)}
\]