How many pairs (a, b) of positive integers are there such that a $\le$ b and $ab=4^{2017}$?

The correct option is 2018.

To find the number of pairs $(a,b)$ of positive integers such that $a\le b$ and $ab=4^{2017}$, we can start by simplifying the expression for the product.

Since $4=2^2$, we can rewrite the equation as follows:

$$ab={\left(2^2\right)}^{2017}=2^{2\times 2017}=2^{4034}$$

Because the product $ab$ is a power of 2, both $a$ and $b$ must be powers of 2. Let us express $a$ and $b$ as:

$$a=2^x\text{and}b=2^y$$

where $x$ and $y$ are non-negative integers. Substituting these into the product equation gives:

$$2^x\times 2^y=2^{x+y}=2^{4034}$$

This implies that the exponents must satisfy the linear relation:

$$x+y=4034$$

We are given the condition $a\le b$. In terms of our exponents, this inequality translates to:

$$2^x\le 2^y\Rightarrow x\le y$$

Since $y=4034-x$, we can substitute this expression into the inequality:

$$x\le 4034-x$$

Adding $x$ to both sides:

$$2x\le 4034$$

Dividing both sides by 2 yields:

$$x\le 2017$$

Since $a$ is a positive integer, the exponent $x$ must be a non-negative integer ($x\ge 0$). Thus, the possible values for $x$ are the integers in the range:

$$0\le x\le 2017$$

The total number of such integers is calculated as:

$$2017-0+1=2018$$

Each chosen value of $x$ uniquely determines a corresponding value of $y$ (where $y=4034-x$ and $y\ge x$), resulting in a unique pair $(a,b)$.

Therefore, there are exactly 2018 pairs of positive integers that satisfy the given conditions.