We rewrite the given equation as
$\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}+\dfrac{1}{s}+\dfrac{1}{t}=\dfrac{1}{5}$
The number of five positive integers $(p,\,q,\,r,\,s,\,t)$ which satisfy the given relation and for which $p\ne q$ is even, because for if $(p,\,q,\,r,\,s,\,t)$ is a solution, then so is $(q,\,p,\,r,\,s,\,t)$ which is distinct from $(p,\,q,\,r,\,s,\,t)$.
Similarly the number of five positive integers which satisfy the equation and for which $r\ne s$ is also even.
Hence, it sufices to count only those five positive integers $(p,\,q,\,r,\,s,\,t)$ for which $p=q$ and $r=s$. Thus the equation reduces to
$\dfrac{2}{p}+\dfrac{2}{r}+\dfrac{1}{t}=\dfrac{1}{5}$
Here again $(p,\,p,\,r,\,r,\,t)$ for which $p\ne r$ is even.
Thus, it suffices to consider the equation
$\dfrac{4}{p}+\dfrac{1}{t}=\dfrac{1}{5}$
and show that the number of pairs $(p,\,t)$ satisfying this equation is odd.
This reduces to
$pt=20t+5p$
$(p-20)(t-5)=100$
Observe that
$100=1\times 100=2\times 50=4\times 25=5\times 20=10\times 10=20\times 5=25\times 4=50\times 2=100\times 1$
Note that no factorization of 100 as product of two negative numbers yield a positive pairs $(p,\,t)$.
Hence we get these 9 solutions. This proves that the total number of five positive integers $(p,\,q,\,r,\,s,\,t)$ satifying the given equation is odd.