Which of the numbers has the largest number of divisors? 
Which of the numbers $1,2,\ldots,1983$ has the largest number of divisors? 

Firstly notice that each number less than or equal to $1983$ has at most $4$ different prime divisors since $2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 > 1983$. We then analyze each of the cases of the amount of prime divisors.
Case 1: $1$ prime divisor 
In this case there are only two divisors. 
Case 2: Here we can have $2^a \cdot 3^b$. The set of all such numbers is $\{2^1 \cdot 3^1,2^1 \cdot 3^2, 2^1 \cdot 3^3, 2^1 \cdot 3^4, 2^1 \cdot 3^5, 2^1 \cdot 3^6, 2^2 \cdot 3^1, 2^2 \cdot 3^2, 2^2 \cdot 3^3, 2^2 \cdot 3^4, 2^2 \cdot 3^5, 2^3 \cdot 3^1, 2^3 \cdot 3^2, 2^3 \cdot 3^3, 2^3 \cdot 3^4, 2^3 \cdot 3^5, 2^4 \cdot 3^1, 2^4 \cdot 3^2, 2^4 \cdot 3^3, 2^4 \cdot 3^4, 2^5 \cdot 3^1, 2^5 \cdot 3^2, 2^5 \cdot 3^3, 2^6 \cdot 3^1, 2^6 \cdot 3^2, 2^6 \cdot 3^3, 2^7 \cdot 3^1, 2^7 \cdot 3^2, 2^8 \cdot 3^1, 2^9 \cdot 3^1\}$. Thus the maximum number of divisors in this case occurs for $2^6 \cdot 3^3$ which is $28$.
The other cases are similar but there are many numbers in the other sets. Is there a more efficient way of going about solving this? 

If $n=p_1^{e_1}\cdots p_r^{e_r}$, then the number of divisors $d(n)$ of $n$ is given by
$$
d(n)=(e_1+1)(e_2+1)\ldots (e_r+1).
$$
We can have at most four different prime divisors, as you said,because otherwise $n>2\cdot 3\cdot5\cdot 7\cdot 11=2310$. Also, we may assume that we have the smallest primes, i.e., $n=2^{e_1}3^{e_2}5^{e_3}7^{e_4}$.
So your method works well, and we obtain, after only cheking a few cases,
$$
d(1680)=d(2^4\cdot 3\cdot 5\cdot 7)=5\cdot 2\cdot 2\cdot 2=40
$$
as the maximum.
