Since $3m^{3}$ is a multiple of 3, then $5n^{5}$ is a multiple of 3. Since 5 is not a multiple of 3 and 3 is a prime number, then $n^{5}$ is a multiple of 3. Since $n^{5}$ is a multiple of 3 and 3 is a prime number, then $n$ is a multiple of 3, which means that $5n^{5}$ includes at least 5 factors of 3. Since $5n^{5}$ includes at least 5 factors of 3, then $3m^{3}$ includes at least 5 factors of 3, which means that $m^{3}$ is a multiple of 3, which means that $m$ is a multiple of 3. Using a similar analysis, both $m$ and $n$ must be multiples of 5. Therefore, we can write $m=3^{a}5^{b}s$ for some positive integers $a, b$ and $s$ and we can write $n=3^{c}5^{d}t$ for some positive integers $c, d$ and $t$, where neither $s$ nor $t$ is a multiple of 3 or 5. From the given equation, $3m^{3}=5n^{5}$, $3(3^{a}5^{b}s)^{3}=5(3^{c}5^{d}t)^{5}$, $3 \times 3^{3a}5^{3b}s^{3}=5 \times 3^{5c}5^{5d}t^{5}$, $3^{3a+1}5^{3b}s^{3}=3^{5c}5^{5d+1}t^{5}$. Since $s$ and $t$ are not multiples of 3 or 5, we must have $3^{3a+1}=3^{5c}$ and $5^{3b}=5^{5d+1}$ and $s^{3}=t^{5}$. Since $s$ and $t$ are positive and $m$ and $n$ are to be as small as possible, we can set $s=t=1$, which satisfy $s^{3}=t^{5}$. Since $3^{3a+1}=3^{5c}$ and $5^{3b}=5^{5d+1}$, then $3a+1=5c$ and $3b=5d+1$. Since $m$ and $n$ are to be as small as possible, we want to find the smallest positive integers $a, b, c, d$ for which $3a+1=5c$ and $3b=5d+1$. Neither $a=1$ nor $a=2$ gives a value for $3a+1$ that is a multiple of 5, but $a=3$ gives $c=2$. Similarly, $b=1$ does not give a value of $3b$ that equals $5d+1$ for any positive integer $d$, but $b=2$ gives $d=1$. Therefore, the smallest possible values of $m$ and $n$ are $m=3^{3}5^{2}=675$ and $n=3^{2}5^{1}=45$, which gives $m+n=720$.