Note: A solution was given many hours ago by Bill Cook, under the assumption that what was asked for is the number of unordered pairs. This was then modified to deal with ordered pairs, and then, for reasons not clear to me, deleted.
We want to count the number of ordered pairs $(i,j)$ such that $i$ and $j$ are positive integers and $ij=N$. Let $X$ be this collection of ordered pairs, and let $Y$ be the set of positive divisors of $N$. Then the number of ordered pairs is the same as the number of positive divisors. This is essentially obvious, so the next paragraph need not be read!
To be formal about it, define the map $\varphi\colon X\to Y$ by $\varphi(i,j)=i$. Then the map $\varphi$ is a bijection. This is not hard to verify. For example, to show that $\varphi$ is onto, let $i$ be any element of $Y$.
Let $j=N/i$. Then $\varphi(i,j)=i$.
Our count of the number of divisors uses the Fundamental Theorem of Arithmetic, aka the Unique Factorization Theorem: Every positive integer has, apart from the order of the terms, a unique representation as a product of prime powers. (The positive integer $1$ is represented as the empty product.)
We show how we can build divisors of $N$ step by step. Imagine the following cafeteria line. First there is a tray with $a$ $2$'s. Next comes a tray with $b$ $3$'s, and then a tray with $c$ $5$'s, and so on, with finally $k+1$ $p$'s. It is OK if some of the trays are empty, for example if $a=0$.
We build a divisor $d$ of $N$ as follows. Go down the cafeteria line. First choose the number of $2$'s that $d$ will "have." More formally, we are choosing the exponent of $2$ in the prime power factorization of $d$. There are $a+1$ possible choices, namely $0,1,\dots,a$.
Continue down the cafeteria line. For every decision about the number of $2$'s, there are $b+1$ ways to decide on the number of $3$'s that $d$ will have. Thus there are $(a+1)(b+1)$ ways to decide on the number of $2$'s and the number of $3$'s.
Continue. For every decision about the number of $2$'s and the number of $3$'s, there are $c+1$ ways to decide on the number of $5$'s that $d$ will have, and so on. Continue. We conclude that the number of positive divisors of $N$, and hence the number of ordered pairs, is equal to
$$(a+1)(b+1)(c+1)\cdots (k+1).$$
Comment $1$: The notation that uses the alphabet in increasing order, as in the current post, used to be standard. That is how Euler wrote. Nowadays indices are ordinarily used, with in my opinion a loss of clarity. With indices, the result would be stated as follows. Suppose that
$$N=p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}=\prod_{i=1}^k p_i^{e_i},$$
where the $p_i$ are distinct primes, and the $e_i$ are positive (or even non-negative) integers. Then the number $d(N)$ of positive divisors of $N$ is given by
$$d(N)=(e_1+1)(e_2+1)\cdots(e_k+1)=\prod_{i=1}^k (e_i+1).$$
Comment $2$: There is a slightly different point of view that is very useful in the long run. For any positive integer $n$, let $d(n)$ be the number of positive divisors of $n$. It can be shown that if $a$ and $b$ are relatively prime (meaning that they have no common positive divisor greater than $1$), then
$d(a,b)=d(a)d(a)$. [A function $f$ defined on the positive integers is called multiplicative if $f(a,b)=f(a)f(b)$ whenever $a$ and $b$ are relatively prime.] Since the function $d(n)$ is multiplicative, if we know $d(p^e)$ for any prime power $p^e$, we know $d(n)$ for any positive integer $n$. But it is clear that $d(p^e)=e+1$.
Let $n \in \mathbb{N}$ and consider its unique prime factorization (fundamental theorem of arithmetic)
$$n = p_1^{a_1} p_2^{a_2} \cdots p_m^{a_m}.$$
How many factors does $n$ have?
Well, a basic counting argument will yield
$$(a_1+1)(a_2+1) \cdots (a_m+1).$$
So, try to come up with relatively small prime factorization that yield 12 here. The integers that you found work.
As an example, if we have $12 = 2^2*3$ then 12 has 6 factors, namely
$$2^0*3^0 = 1, 2^1*3^0 = 2, 2^2*3^0 = 4, 2^0*3^1 = 3, 2^1*3^1 = 6, 2^2*3^1 =12.$$
Further, to count the number of divisors (or factors) that an integer has, it is common to use the arithmetic function $\sigma_0(n)$. The function is multiplicative which means that if we have integers $k, l$ such that $k$ and $l$ are relatively prime (no common prime factors) then $\sigma_0(kl) = \sigma_0(k) \sigma_0(l) $. The upshot here is that powers of distinct primes are relatively prime and so if you know the prime factorization for an integer $n$ then you can calculate $\sigma_0(n)$ relatively fast.
For more on this, see the properties of the divisor function here: https://en.m.wikipedia.org/wiki/Divisor_function
I hope this helps!
Best Answer
Hint: if the prime factorization of $\,n\in\Bbb N\,$ is
$$n=\prod_{k=1}^rp_k^{a_k}$$
then the number of divisors of $\,n\,$ is
$$\prod_{k=1}^r(a_k+1)$$