Find all the positive integers $n$ so that $n$ has the same number of digits as its number of different prime factors and the sum of these different prime factors is equal to the sum of exponents of all these primes in factorization of $n$.
To solve the problem, we need to find all positive integers \( n \) such that:
1. \( n \) has the same number of digits as its number of different prime factors.
2. The sum of these different prime factors is equal to the sum of the exponents of all these primes in the factorization of \( n \).

Let \( k \) be the number of digits of \( n \). Then, we have:
\[ n = \prod_{i=1}^k p_i^{e_i} \]
where \( p_i \) are distinct prime factors and \( e_i \) are their respective exponents.

### Step 1: Establishing the bounds for \( n \)
Since \( n \) has \( k \) digits, we have:
\[ 10^{k-1} \leq n < 10^k \]

### Step 2: Relating \( n \) to its prime factors and exponents
Given the conditions, we have:
\[ \sum_{i=1}^k p_i = \sum_{i=1}^k e_i \]

### Step 3: Analyzing the inequality
We also know:
\[ n > \prod_{i=1}^k 2^{e_i} = 2^{\sum_{i=1}^k e_i} = 2^{\sum_{i=1}^k p_i} \]
Thus:
\[ 10^{k-1} \leq n < 10^k \]
\[ 10^k > 2^{\sum_{i=1}^k p_i} \]

### Step 4: Simplifying the inequality
Since \( \sum_{i=1}^k p_i \geq k \cdot 2 \) (because the smallest prime is 2), we have:
\[ 10^k > 2^{k \cdot 2} = 2^{2k} \]
Taking the logarithm base 10 on both sides:
\[ k \log_{10}(10) > 2k \log_{10}(2) \]
\[ k > 2k \log_{10}(2) \]
\[ k > 2k \cdot 0.3010 \]
\[ k > 0.602k \]
\[ k (1 - 0.602) > 0 \]
\[ k \cdot 0.398 > 0 \]
Thus, \( k \leq 2 \).

### Step 5: Case \( k = 2 \)
If \( k = 2 \), then \( n \) has 2 digits. We need to find \( n \) such that:
\[ n = p_1^{e_1} p_2^{e_2} \]
with \( p_1 + p_2 = e_1 + e_2 \).

Since \( n \) is a two-digit number:
\[ 10 \leq n < 100 \]
\[ 10^2 > 2^{p_1 + p_2} \]
\[ 100 > 2^{p_1 + p_2} \]
\[ p_1 + p_2 \leq 6 \]

The possible pairs \((p_1, p_2)\) are \((2, 3)\) since they are the smallest primes. Thus:
\[ p_1 + p_2 = 2 + 3 = 5 \]
\[ e_1 + e_2 = 5 \]

Possible values of \( n \) are:
\[ n = 2^4 \cdot 3^1 = 48 \]
\[ n = 2^3 \cdot 3^2 = 72 \]

### Step 6: Case \( k = 1 \)
If \( k = 1 \), then \( n \) has 1 digit. We need to find \( n \) such that:
\[ n = p^e \]
with \( p = e \).

The only single-digit number that is a power of a prime is:
\[ n = 2^2 = 4 \]

### Conclusion
The solutions are \( n = 4, 48, 72 \).

The final answer is \(\boxed{4, 48, 72}\)