A positive integer is equal to the sum of the squares of its four smallest positive divisors. What is the largest prime that divides this positive integer?
1. Let the number be \( n \), and let its four smallest positive divisors be \( 1, a, b, c \). Then we have:
   \[
   n = 1 + a^2 + b^2 + c^2
   \]
2. Since \( 1 \) is always the smallest divisor, we need to consider the next smallest divisor \( a \). If \( a \neq 2 \), then \( a, b, c, n \) are all odd. This would make the left-hand side (LHS) odd and the right-hand side (RHS) even, which is a contradiction. Therefore, \( a = 2 \).

3. Substituting \( a = 2 \) into the equation, we get:
   \[
   n = 1 + 2^2 + b^2 + c^2 = 5 + b^2 + c^2
   \]

4. Next, we consider the condition \( 4 \mid n \). If \( 4 \mid n \), then one of \( b \) or \( c \) must be equal to \( 4 \). However, if \( b = 4 \) or \( c = 4 \), then:
   \[
   n = 5 + 4^2 + c^2 = 5 + 16 + c^2 = 21 + c^2
   \]
   or
   \[
   n = 5 + b^2 + 4^2 = 5 + b^2 + 16 = 21 + b^2
   \]
   In either case, \( 21 + c^2 \) or \( 21 + b^2 \) must be divisible by 4, which is not possible since \( 21 \equiv 1 \pmod{4} \). Therefore, \( b \neq 4 \).

5. Since \( b \neq 4 \), \( b \) must be a prime number and odd. We take \( b \) modulo 2:
   \[
   0 \equiv c^2 \pmod{2}
   \]
   This implies \( c \) must be even. Since \( c \) is even and greater than 2, it must be \( 2b \).

6. Substituting \( c = 2b \) into the equation, we get:
   \[
   n = 5 + b^2 + (2b)^2 = 5 + b^2 + 4b^2 = 5 + 5b^2 = 5(1 + b^2)
   \]

7. To find \( b \), we need \( 1 + b^2 \) to be an integer. Let \( b = 5 \):
   \[
   n = 5(1 + 5^2) = 5(1 + 25) = 5 \cdot 26 = 130
   \]

8. The prime factorization of \( 130 \) is:
   \[
   130 = 2 \times 5 \times 13
   \]
   The largest prime divisor is \( 13 \).

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