First Solution: Let $s$ denote the minimum number of cells in a dinosaur; the number this year is $s = 2007$.

Claim: The maximum number of cells in a primitive dinosaur is $4(s - 1) + 1 = 8025$.

First, a primitive dinosaur can contain up to $4(s - 1) + 1$ cells. To see this, consider a dinosaur in the form of a cross consisting of a central cell and four arms with $s - 1$ cells apiece. No connected figure with at least $s$ cells can be removed without disconnecting the dinosaur.

The proof that no dinosaur with at least $4(s - 1) + 2$ cells is primitive relies on the following result.

Lemma Let $D$ be a dinosaur having at least $4(s - 1) + 2$ cells, and let $R$ (red) and $B$ (black) be two complementary animals in $D$, i.e., $R \cap B = \emptyset$ and $R \cup B = D$. Suppose $|R| \le s - 1$. Then $R$ can be augmented to produce animals $\tilde{R} \supset R$ and $\tilde{B} = D \setminus \tilde{R}$ such that at least one of the following holds:

(i) $|\tilde{R}| \ge s$ and $|\tilde{B}| \ge s$,

(ii) $|\tilde{R}| = |R| + 1$,

(iii) $|R| < |\tilde{R}| \le s - 1$.

Proof: If there is a black cell adjacent to $R$ that can be made red without disconnecting $B$, then (ii) holds. Otherwise, there is a black cell $c$ adjacent to $R$ whose removal disconnects $B$. Of the squares adjacent to $c$, at least one is red, and at least one is black, otherwise $B$ would be disconnected. Then there are at most three resulting components $C_1, C_2, C_3$ of $B$ after the removal of $c$. Without loss of generality, $C_3$ is the largest of the remaining components. (Note that $C_1$ or $C_2$ may be empty.) Now $C_3$ has at least $\lceil(3s - 2)/3\rceil = s$ cells. Let $\tilde{B} = C_3$. Then $|\tilde{R}| = |R| + |C_1| + |C_2| + 1$. If $|\tilde{B}| \le 3s - 2$, then $|\tilde{R}| \ge s$ and (i) holds. If $|\tilde{B}| \ge 3s - 1$ then either (ii) or (iii) holds, depending on whether $|\tilde{R}| \ge s$ or not.

Starting with $|R| = 1$, repeatedly apply the Lemma. Because in alternatives (ii) and (iii) $|R|$ increases but remains less than $s$, alternative (i) eventually must occur. This shows that no dinosaur with at least $4(s - 1) + 2$ cells is primitive.

Second Solution: (Based on Andrew Geng's solution) Let $s = 2007$. We claim that the answer is $4s - 3 = 8025$.

Consider a graph with the cells as the vertices and whose edges connect adjacent cells. Let $T$ be a spanning tree in this graph. By removing any vertex of $T$, we obtain at most four connected components, which we call the limbs of the vertex. Limbs with at least $s$ vertices are called big. Suppose that every vertex of $T$ contains a big limb, then consider a walk on $T$ starting from an arbitrary vertex and always moving along the edge towards a big limb. Since $T$ is a finite tree, this walk must traverse back on some edge at some point. Then the two connected components of $T$ made by deleting this edge are both big, so they both contain at least $s$ vertices, which means that the dinosaur is not primitive. It follows that a primitive dinosaur contains some vertex with no big limbs. By removing this vertex, we get at most four connected components with at most $s - 1$ vertices each. This not only shows that a primitive dinosaur has at most $4s - 3$ cells, but also shows that any such dinosaur consists of four limbs of $s-1$ cells each connected to a central cell. It is easy to see that such a dinosaur indeed exists.