What is the largest positive integer $n$ such that $$\frac{a^2}{\frac{b}{29} + \frac{c}{31}}+\frac{b^2}{\frac{c}{29} + \frac{a}{31}}+\frac{c^2}{\frac{a}{29} + \frac{b}{31}} \ge n(a+b+c)$$holds for all positive real numbers $a,b,c$.
1. We start with the given inequality:
   \[
   \frac{a^2}{\frac{b}{29} + \frac{c}{31}} + \frac{b^2}{\frac{c}{29} + \frac{a}{31}} + \frac{c^2}{\frac{a}{29} + \frac{b}{31}} \ge n(a+b+c)
   \]
   for all positive real numbers \(a, b, c\).

2. To simplify the left-hand side, we use the Titu's Lemma (or Engel's form of the Cauchy-Schwarz inequality), which states:
   \[
   \frac{x_1^2}{y_1} + \frac{x_2^2}{y_2} + \cdots + \frac{x_n^2}{y_n} \ge \frac{(x_1 + x_2 + \cdots + x_n)^2}{y_1 + y_2 + \cdots + y_n}
   \]
   Here, we set \(x_1 = a, x_2 = b, x_3 = c\) and \(y_1 = \frac{b}{29} + \frac{c}{31}, y_2 = \frac{c}{29} + \frac{a}{31}, y_3 = \frac{a}{29} + \frac{b}{31}\).

3. Applying Titu's Lemma, we get:
   \[
   \frac{a^2}{\frac{b}{29} + \frac{c}{31}} + \frac{b^2}{\frac{c}{29} + \frac{a}{31}} + \frac{c^2}{\frac{a}{29} + \frac{b}{31}} \ge \frac{(a+b+c)^2}{\left(\frac{b}{29} + \frac{c}{31}\right) + \left(\frac{c}{29} + \frac{a}{31}\right) + \left(\frac{a}{29} + \frac{b}{31}\right)}
   \]

4. Simplify the denominator:
   \[
   \left(\frac{b}{29} + \frac{c}{31}\right) + \left(\frac{c}{29} + \frac{a}{31}\right) + \left(\frac{a}{29} + \frac{b}{31}\right) = \frac{b+c}{29} + \frac{c+a}{31} + \frac{c+a}{29} + \frac{a+b}{31} = \left(\frac{1}{29} + \frac{1}{31}\right)(a+b+c)
   \]

5. Calculate the sum of the fractions:
   \[
   \frac{1}{29} + \frac{1}{31} = \frac{31 + 29}{29 \cdot 31} = \frac{60}{899}
   \]

6. Substitute back into the inequality:
   \[
   \frac{(a+b+c)^2}{\left(\frac{60}{899}\right)(a+b+c)} = \frac{(a+b+c)^2}{\frac{60}{899}(a+b+c)} = \frac{899}{60}(a+b+c)
   \]

7. Therefore, we have:
   \[
   \frac{a^2}{\frac{b}{29} + \frac{c}{31}} + \frac{b^2}{\frac{c}{29} + \frac{a}{31}} + \frac{c^2}{\frac{a}{29} + \frac{b}{31}} \ge \frac{899}{60}(a+b+c)
   \]

8. To find the largest integer \(n\) such that:
   \[
   \frac{899}{60} \ge n
   \]
   we calculate:
   \[
   \frac{899}{60} \approx 14.9833
   \]
   The largest integer less than or equal to 14.9833 is 14.

9. Therefore, the largest integer \(n\) is 14.

The final answer is \(\boxed{14}\).