The answer is $2146$.

Note that $2188 = 4 \times 547$, where $547$ is a prime. So $2188 = 40_{(547)}$ (base $547$ representation). Let $k = \overline{ab}_{(547)}$. If $k$ is not divisible by $547$, then $b > 0$. By Lucas' theorem,
$$\binom{2188}{k} = \binom{40_{(547)}}{\overline{ab}_{(547)}} \equiv \binom{4}{a}\binom{0}{b} = 0 \pmod{547}.$$
This shows $\binom{2188}{k}$ is divisible by $547$. If $547 \mid k$, then $b=0$ and $a \le 4$. By Lucas' theorem,
$$\binom{2188}{k} = \binom{40_{(547)}}{\overline{a0}_{(547)}} \equiv \binom{4}{a}\binom{0}{0} = \binom{4}{a} \not\equiv 0 \pmod{547}.$$
This shows $\binom{2188}{k}$ is not divisible by $547$.

Next, let $N$ be the highest power of $2$ dividing $\binom{2188}{k}$. Using binary representation, we have $2188 = 100010001100_{(2)}$. By Kummer's theorem, $N$ is the number of carries when $k$ is added to $2188-k$ in base $2$.

Firstly, $\binom{2188}{k}$ is odd if and only if $N = 0$. This holds if and only if $a_j$ is $0$ whenever the corresponding digits of $2188$ in binary representation are $0$. In other words, $k = \overline{a000b000cd00}_{(2)}$. There are $2^4 = 16$ such numbers.

Secondly, $\binom{2188}{k}$ is even and is not divisible by $4$ if and only if $N = 1$. This holds if and only if $k$ has the form $\overline{0100a000bc00}_{(2)}$, $\overline{a0000100bc00}_{(2)}$, $\overline{a000b000c010}_{(2)}$. There are $2^3 \times 3 = 24$ such numbers.

Finally, $\binom{2188}{0} = \binom{2188}{2188} = 1$ are odd, while $4 \mid \binom{2188}{547}, \binom{2188}{1094}, \binom{2188}{1641}$ since $547 = \overline{001000100011}_{(2)}$, $1094 = \overline{010001000110}_{(2)}$, $1641 = \overline{011001101001}_{(2)}$. Therefore, the final answer is
$$2189 - 5 - 16 - 24 + 2 = 2146.$$