Find all prime numbers $p$, for which there exist $x, y \in \mathbb{Q}^+$ and $n \in \mathbb{N}$, satisfying
$x+y+\frac{p}{x}+\frac{p}{y}=3n$.
1. We start with the given equation:
   \[
   x + y + \frac{p}{x} + \frac{p}{y} = 3n
   \]
   where \( p \) is a prime number, \( x, y \in \mathbb{Q}^+ \) (positive rational numbers), and \( n \in \mathbb{N} \) (natural numbers).

2. Suppose \( x = \frac{a}{b} \) and \( y = \frac{c}{d} \) where \( a, b, c, d \in \mathbb{N} \) and \(\gcd(a, b) = \gcd(c, d) = 1\). Substituting these into the equation, we get:
   \[
   \frac{a}{b} + \frac{c}{d} + \frac{p \cdot b}{a} + \frac{p \cdot d}{c} = 3n
   \]
   Multiplying through by \( b \cdot d \cdot a \cdot c \) to clear the denominators, we obtain:
   \[
   (ad + bc)(ac + pbd) = 3n \cdot abcd
   \]
   This implies:
   \[
   abcd \mid (ad + bc)(ac + pbd)
   \]

3. Let \( h = \gcd(b, d) \). Since \( h \mid b \) and \( h \mid d \), and \(\gcd(h, ac) = 1\), we have \( h = 1 \). Therefore, \( b \mid c \) and \( d \mid a \). We can write \( a = ds \) and \( c = bt \) for some \( s, t \in \mathbb{N} \).

4. Substituting \( a = ds \) and \( c = bt \) into the equation, we get:
   \[
   \frac{(d^2s + b^2t)(st + p)}{bdst}
   \]
   Simplifying, we have:
   \[
   \frac{(d^2s + b^2t)(st + p)}{bdst}
   \]

5. Let \( g = \gcd(s, t) \). Since \( g \mid p \), we consider two cases: \( g = 1 \) and \( g = p \).

6. **Case 1: \( g = 1 \)**
   - Since \(\gcd(s, bt) = 1\), we have \( s \mid p \) and \( t \mid p \). Therefore, \( s \) and \( t \) can be either 1 or \( p \).

   - **Sub-case 1.1: \( (s, t) = (1, 1) \)**
     \[
     F = \frac{(b^2 + d^2)(p + 1)}{bd}
     \]
     Since \( 3 \nmid p + 1 \), we must have \( 3 \mid b^2 + d^2 \). However, if \( 3 \mid b^2 + d^2 \), then \( 3 \mid b \) and \( 3 \mid d \), which contradicts \(\gcd(b, d) = 1\).

   - **Sub-case 1.2: \( (s, t) = (p, 1) \)**
     \[
     F = \frac{2(p d^2 + b^2)}{bd}
     \]
     Since \(\gcd(b, dp) = 1\), we have \( b \mid 2 \) and \( d \mid 2 \). In particular, \(\gcd(bd, 3) = 1\) because \( b, d \leq 2 \). So \( p d^2 + b^2 \equiv p + 1 \pmod{3} \), which is not divisible by 3. Thus, we get a contradiction.

7. **Case 2: \( g = p \)**
   - We can take \( s = pu \) and \( t = pv \) where \(\gcd(u, v) = 1\). So we have:
     \[
     F = \frac{(d^2 u + b^2 v)(p uv + 1)}{bduv}
     \]
     Since \(\gcd(u, vb) = 1\), we have \( u \mid 1 \), which means \( u = 1 \). Similarly, \( v \mid 1 \), which means \( v = 1 \). So we have:
     \[
     F = \frac{(d^2 + b^2)(p + 1)}{bd}
     \]
     We can show that this is a contradiction in the same way as Sub-case 1.1.

Therefore, the only possible solution is when \( 3 \mid p + 1 \).

The final answer is \( \boxed{ p } \) such that \( 3 \mid p + 1 \).