Find all pairs of integers $(c, d)$, both greater than 1, such that the following holds:

For any monic polynomial $Q$ of degree $d$ with integer coefficients and for any prime $p > c(2c+1)$, there exists a set $S$ of at most $\big(\tfrac{2c-1}{2c+1}\big)p$ integers, such that
\[\bigcup_{s \in S} \{s,\; Q(s),\; Q(Q(s)),\; Q(Q(Q(s))),\; \dots\}\]
contains a complete residue system modulo $p$ (i.e., intersects with every residue class modulo $p$).
1. **Claim**: The answer is that all pairs \((c, d)\) with \(d \leq c\) work.

2. **Proof for \(d > c\)**:
   - Use Dirichlet's theorem to pick a large prime \(p\) such that \(p \equiv 1 \pmod{d}\).
   - Set \(Q(x) \equiv x^d \pmod{p}\).
   - For large \(p\), we must pick all nonzero residues modulo \(p\) that are not \(d\)-th powers.
   - The number of such residues is at least \((p-1) \cdot \frac{d-1}{d}\).
   - For large \(p\), \((p-1) \cdot \frac{d-1}{d} > \frac{2c-1}{2c+1} p\).
   - Hence, for \(d > c\), the condition fails.

3. **Proof for \(d \leq c\)**:
   - Construct a directed graph (digraph) on \(p\) vertices where each vertex corresponds to a residue class modulo \(p\) and edges are defined by \(x \rightarrow Q(x)\).
   - Each connected component of this digraph is called a sunflower.
   - A sunflower consists of a central cycle with trees leading into some vertices of the cycle.
   - Define a sunflower as:
     - **Dead**: If it is just one central cycle with no trees.
     - **Critical condition**: If no vertex has a depth of at least two.
     - **Healthy**: Otherwise.

4. **Covering Set**:
   - A set \(T\) of vertices covers the garden if for every vertex \(v\), there is a vertex \(w \in T\) such that we can walk from \(w\) to \(v\) along the digraph's edges.
   - Each vertex has outdegree 1 and indegree \(\leq d\).
   - There are at most \(d\) fixed points by Lagrange's theorem on \(Q(x) - x\).

5. **Induction on \(n\)**:
   - **Base Case**: For \(n = 1\), the covering set is the single vertex itself.
   - **Inductive Step**:
     - If there is a dead sunflower with at least 2 vertices, add one vertex to the covering set of the remaining garden.
     - If there is a healthy sunflower, consider a vertex \(v\) with depth at least 2.
     - Let \(x \leq d\) be the number of vertices mapping to \(Q(v)\).
     - Apply the induction hypothesis on the garden obtained by removing these vertices.
     - The covering set for the remaining garden has at most \(\frac{d-1}{d} (n-x) + 1\) vertices.
     - Add back the \(x\) vertices to obtain a covering set for the original garden: \(\frac{d-1}{d} (n-x) + x \leq \frac{d-1}{d} n + 1\).
     - If there is a sunflower in critical condition, induct on the rest of the garden and add back the leaves of that sunflower.
     - The remaining digraph consists only of dead sunflowers with one vertex in their central cycle.
     - There are at most \(d\) such sunflowers, and \(x \leq \frac{d-1}{d} x + 1\) for all \(x \leq d\).

6. **Conclusion**:
   - By the claim, we can pick a covering set of at most \(\frac{d-1}{d} p + 1\) vertices.
   - Since \(d \leq c\) and \(p > c(2c+1)\), we have \(\frac{d-1}{d} p + 1 \leq \frac{c-1}{c} p + 1 \leq \frac{2c-1}{2c+1} p\).
   - This completes the proof.

\(\blacksquare\)

The final answer is all pairs \( \boxed{ (c, d) } \) with \(d \leq c\).