4. Find all real numbers $p$ such that the cubic equation $5 x^{3}-5(p+1) x^{2}+(71 p-1) x+1=66 p$ has three positive integer roots.

 untranslated text remains the same as requested.
4. Solution 1 It is easy to see that $x=1$ is a positive integer root of the original cubic equation.

By synthetic division, the original cubic equation can be reduced to a quadratic equation:
$$
5 x^{2}-5 p x+66 p-1=0 .
$$

The three roots of the original cubic equation are all positive integers $\Leftrightarrow$ the two roots of the quadratic equation $(*)$ are both positive integers.
Let $u, v(u \leqslant v)$ be the two roots of equation $(*)$, then by the relationship between roots and coefficients, we get
$$
\left\{\begin{array}{l}
u+v=p, \\
u v=\frac{1}{5}(66 p-1) .
\end{array}\right.
$$

Substituting (1) into (2), we get $5 u v=66(u+v)-1$.
It is known that $u, v$ cannot be divisible by $2,3,11$.
From (3), we get $u=\frac{66 v-1}{5 v-66}$.
Since $u, v$ are both positive integers, from (4), we know $u>\frac{66}{5}$, i.e., $u \geqslant 14$.
Also, $2 \times u$ and $3 \times u$, so $u \geqslant 17$. By $v \geqslant u$ and (4), we get $\frac{66 u-1}{5 u-66} \geqslant u$, i.e., $5 u^{2}-132 u+1 \leqslant 0$.
Thus, $u \leqslant \frac{66+\sqrt{66^{2}-5}}{5}<\frac{132}{5}$, therefore $17 \leqslant u \leqslant 26$. Furthermore, by $2 \chi_{u}$ and $3 X_{u}$, $u$ can only take $17,19,23,25$.
When $u=17$, from (4), we get $v=\frac{1121}{19}=59$.
When $u=19$, from (4), we get $v=\frac{1253}{29}$ which is not a positive integer, so it should be discarded.
When $u=23$, from (4), we get $v=\frac{1517}{49}$ which is not a positive integer, so it should be discarded.
When $u=25$, from (4), we get $v=\frac{1649}{59}$ which is not a positive integer, so it should be discarded.
Therefore, only when $p=u+v=76$, the two roots of equation (*) are both positive integers, and the three roots of the original equation are all positive integers.
Solution 2 Using the notation in Solution 1, from (1)(2), we get $5 u v=66(u+v)-1$,
i.e., $5^{2} u v=5 \times 66(u+v)-5$.
The above equation can be factored into $(5 u-66)(5 v-66)=19 \times 229$.
The above indeterminate equation can be factored into $\left\{\begin{array}{l}5 v-66=229, \\ 5 u-66=19 ;\end{array}\right.$ or $\left\{\begin{array}{l}5 v-66=19 \times 229, \\ 5 u-66=1 .\end{array}\right.$
The latter has no positive integer solutions, the former's solution is $v=59, u=17$.
Therefore, when $p=u+v=17+59=76$, the two roots of equation (*) are both positive integers, i.e., when $p=76$, the three roots of the original cubic equation are all positive integers.
Comment When $p=76$, the two roots of the quadratic equation (*) are 59 and 17; the three positive integer roots of the original cubic equation are 1, 59, and 17.