To solve the problem, we need to factorize the number 4²⁰⁰⁷.
4 can be written as 2², so 4²⁰⁰⁷ can be written as (2²)²⁰⁰⁷, which equals 2⁴⁰⁰⁷.
So, ab = 2⁴⁰⁰⁷.


We need to count the number of pairs (a,b) such that a≤b and ab = 2⁴⁰⁰⁷.
Let's consider the prime factorization of 2⁴⁰⁰⁷:
2⁴⁰⁰⁷ = 2¹ x 2¹ x … x 2¹ (2007 times)


To form a pair (a,b), we need to divide the 2007 factors of 2 into two groups.
Let's say we divide them into two groups of sizes p and q such that p+q=2007.
We can form a pair (a,b) by taking the product of the factors in the two groups:
a = 2¹ x 2¹ x … x 2¹ (p times)
b = 2¹ x 2¹ x … x 2¹ (q times)


Since a≤b, we can assume that p≤q.
So, we need to count the number of ways to choose p from 0 to 2007.
The number of ways to choose p is the same as the number of ways to choose q, so we only need to count up to 1004 (half of 2007).


Therefore, the total number of pairs (a,b) is the sum of the number of pairs for each p from 0 to 1004:
∑(2007 choose p) = 2²⁰⁰⁷
So, there are 2018 pairs (a,b) of positive integers such that a≤b, and ab = 4²⁰⁰⁷.