Solution:

Let us first write the general term for each sequence.

Lemuel's sequence: $2, 9, 16, 23, \ldots$
This is an arithmetic sequence with first term $a_1 = 2$ and common difference $d = 7$.
So the $n$th term is $a_n = 2 + 7(n-1) = 7n - 5$.

Christine's sequence: $3, 7, 11, 15, \ldots$
This is an arithmetic sequence with first term $b_1 = 3$ and common difference $d = 4$.
So the $m$th term is $b_m = 3 + 4(m-1) = 4m - 1$.

We are to find how many numbers appear in both sequences among the first 2010 terms of each.

A number is in both sequences if $7n - 5 = 4m - 1$ for some integers $n, m$ with $1 \leq n \leq 2010$ and $1 \leq m \leq 2010$.

So $7n - 5 = 4m - 1 \implies 7n - 4m = 4$.

We want integer solutions $(n, m)$ with $1 \leq n \leq 2010$, $1 \leq m \leq 2010$.

Let us solve $7n - 4m = 4$ for integers $n, m$.

$7n - 4m = 4 \implies 7n = 4m + 4 \implies n = \frac{4m + 4}{7}$.

We need $n$ to be integer, so $4m + 4 \equiv 0 \pmod{7}$.

$4m + 4 \equiv 0 \pmod{7} \implies 4m \equiv -4 \pmod{7} \implies 4m \equiv 3 \pmod{7}$ (since $-4 \equiv 3 \pmod{7}$).

Now, $4$ and $7$ are coprime, so $4$ has an inverse modulo $7$.

The inverse of $4$ modulo $7$ is $2$, since $4 \times 2 = 8 \equiv 1 \pmod{7}$.

So $m \equiv 2 \times 3 \pmod{7} \implies m \equiv 6 \pmod{7}$.

So $m = 7k + 6$ for integer $k \geq 0$.

Now, $1 \leq m \leq 2010$.

So $7k + 6 \leq 2010 \implies 7k \leq 2004 \implies k \leq 286.285...$

So $k$ ranges from $0$ to $286$ (inclusive), so $k = 0, 1, 2, \ldots, 286$.

Thus, there are $287$ possible values of $m$.

Now, for each $m = 7k + 6$, $n = \frac{4m + 4}{7} = \frac{4(7k + 6) + 4}{7} = \frac{28k + 24 + 4}{7} = \frac{28k + 28}{7} = 4k + 4$.

We need $1 \leq n \leq 2010$.

For $k = 0$, $n = 4$.
For $k = 286$, $n = 4 \times 286 + 4 = 1144 + 4 = 1148$.

So $n$ ranges from $4$ to $1148$ in steps of $4$.

But since $k$ runs from $0$ to $286$, there are $287$ values.

Therefore, the answer is $\boxed{287}$.