1. We need the constant term in $\left(2x^3+\frac{3}{x^k}\right)^{12}$ for positive integers $k$.

2. General term in the expansion: $T_{r+1}=\binom{12}{r}(2x^3)^{12-r}\left(\frac{3}{x^k}\right)^r$ where $r=0,1,2,\dots,12$.

3. Simplify the power of $x$: $T_{r+1}=\binom{12}{r}2^{12-r}3^r\,x^{3(12-r)-kr}$ So exponent of $x$ is $36-3r-kr=36-r(k+3).$

4. For the term to be constant, exponent of $x$ must be zero: $36-r(k+3)=0$ $r(k+3)=36.$

Thus $r$ must be a divisor of $36$, and $k=\frac{36}{r}-3.$ Since $k$ is a positive integer, $\frac{36}{r}-3>0 \implies \frac{36}{r}>3 \implies r<12.$ Also $r$ must be an integer between $0$ and $12$.

5. List divisors $r$ of $36$ with $1\le r<12$: $r=1,2,3,4,6,9.$ These give:

• $r=1 \Rightarrow k=36-3=33$
• $r=2 \Rightarrow k=18-3=15$
• $r=3 \Rightarrow k=12-3=9$
• $r=4 \Rightarrow k=9-3=6$
• $r=6 \Rightarrow k=6-3=3$
• $r=9 \Rightarrow k=4-3=1$

So possible positive integers $k$ are $1,3,6,9,15,33.$ There are $6$ such values.

6. The question says this number is $2l$, where $l$ is an odd integer. Since the number is $6$, $2l=6 \Rightarrow l=3.$

7. Since $3$ is odd, it satisfies the condition.

Therefore, $\boxed{3}$

8. Comparison with stored answer: Stored correct answer is $2$, but the derived value is $3$. Hence the stored answer appears incorrect.