For a positive integer $n$, $S(n)$ denotes the sum of its digits and $U(n)$ its unit digit. Determine all positive integers $n$ with the property that
\[n = S(n) + U(n)^2.\]
To solve the problem, we need to find all positive integers \( n \) such that:
\[ n = S(n) + U(n)^2, \]
where \( S(n) \) denotes the sum of the digits of \( n \) and \( U(n) \) denotes the unit digit of \( n \).

1. **Determine the number of digits \( t \) in \( n \):**
   - If \( n \) has \( t \) digits, then the maximum value of \( S(n) \) is \( 9t \) (since each digit can be at most 9).
   - The unit digit \( U(n) \) can be at most 9, so \( U(n)^2 \leq 81 \).
   - Therefore, we have:
     \[
     n = S(n) + U(n)^2 \leq 9t + 81.
     \]
   - This implies:
     \[
     n \leq 9t + 81.
     \]

2. **Analyze the possible values of \( t \):**
   - For \( t = 1 \):
     - If \( n \) has only one digit, then \( n = S(n) = U(n) \).
     - The equation becomes:
       \[
       n = n + n^2,
       \]
       which is impossible for any positive integer \( n \).

   - For \( t = 2 \):
     - Let \( n = 10a + b \), where \( a \) and \( b \) are the digits of \( n \) (with \( b \) being the unit digit).
     - The equation becomes:
       \[
       10a + b = a + b + b^2.
       \]
     - Simplifying, we get:
       \[
       9a = b^2.
       \]
     - Since \( a \) and \( b \) are digits, \( b^2 \) must be a multiple of 9. The possible values of \( b \) are 0, 3, 6, and 9.
       - If \( b = 0 \), then \( 9a = 0 \), so \( a = 0 \), and \( n = 0 \), which is not a positive integer.
       - If \( b = 3 \), then \( 9a = 9 \), so \( a = 1 \), and \( n = 10 \cdot 1 + 3 = 13 \).
       - If \( b = 6 \), then \( 9a = 36 \), so \( a = 4 \), and \( n = 10 \cdot 4 + 6 = 46 \).
       - If \( b = 9 \), then \( 9a = 81 \), so \( a = 9 \), and \( n = 10 \cdot 9 + 9 = 99 \).

   - For \( t = 3 \):
     - From the inequality \( n \leq 9t + 81 \), we have \( n \leq 108 \).
     - We need to check if there are any three-digit numbers \( n \) that satisfy the equation.
     - Let \( n = 100a + 10b + c \), where \( a, b, c \) are the digits of \( n \) (with \( c \) being the unit digit).
     - The equation becomes:
       \[
       100a + 10b + c = a + b + c + c^2.
       \]
     - Simplifying, we get:
       \[
       99a + 9b = c^2.
       \]
     - Since \( c^2 \) must be a multiple of 9, the possible values of \( c \) are 0, 3, 6, and 9.
       - If \( c = 0 \), then \( 99a + 9b = 0 \), so \( a = 0 \) and \( b = 0 \), and \( n = 0 \), which is not a positive integer.
       - If \( c = 3 \), then \( 99a + 9b = 9 \), which is not possible since \( 99a + 9b \) must be a multiple of 9.
       - If \( c = 6 \), then \( 99a + 9b = 36 \), which is not possible since \( 99a + 9b \) must be a multiple of 9.
       - If \( c = 9 \), then \( 99a + 9b = 81 \), which is not possible since \( 99a + 9b \) must be a multiple of 9.

Therefore, the only solutions are \( n = 13, 46, 99 \).

The final answer is \( \boxed{13, 46, 99} \).