We are tasked with determining how many positive integers $n \leq 2022$ are $inversosimil$, which means $n$ can be expressed with $n$ integers such that both the sum and the product of these integers equal $n$.

To solve this problem, let's first consider the sequence of integers that can satisfy the condition. Suppose we have $n$ integers $a_1, a_2, \ldots, a_n$ such that:

$$a_1 + a_2 + \cdots + a_n = n \quad \text{and} \quad a_1 \times a_2 \times \cdots \times a_n = n$$

For this to hold, one straightforward solution is to set $n-1$ of these integers to be 1, and the last integer to be $n - (n-1) = 1$. This leads us to the sequence of all ones:

$$1, 1, \ldots, 1$$

This sequence sums to $n$ and the product is also 1, which equals $n$ when $n = 1$.

However, to meet the criteria for other values of $n$, let us consider a more practical setup. If we have $n - 1$ ones and one last integer $a$ such that:

$$1 + 1 + \cdots + 1 + a = n$$

This equates to:

$$n - 1 + a = n \quad \Rightarrow \quad a = 1$$

The product is:

$$1 \times 1 \times \cdots \times 1 \times (n - (n-1)) = 1$$

This is also satisfied if all but one integer are 1, and the last $a_n = n$, which corresponds to consideting:

$$1 + 1 + \cdots + 1 + (n-(n-1))= n \quad \text{and} \quad 1 \times \cdots \times 1 \times n = n$$

Thus, integers $n$ are $inverosimil$ specifically when they can also be expressed as sequences of ones and a single additional $n$. This usually occurs with even numbers greater than 2.

Reinterpreting this pattern across all $n$ up to 2022, you can identify that all even numbers $n = 2k$, where $k$ is a positive integer, will fit this description since they allow such balanced sequences of contributing numbers. The even numbers less than or equal to 2022 range from 2 to 2022, inclusively.

Hence, considering the values:

$$2, 4, 6, \ldots, 2022$$

We establish the total number of terms in this sequence by using an arithmetic progression formula where each difference between terms is 2. Finally,

$$\frac{2022 - 2}{2} + 1 = 1010$$

The number of $inverosimil$ integers less than or equal to 2022 is therefore:

$$\boxed{1010}$$