The answer is no: if $\lambda$ is a larger than $\omega^2$ and if $X$ contains $\lambda+\omega$ then it also contains $\lambda+\omega+\omega$.
To see this simply observe that $\lambda+1$ is homeomorphic with $\lambda+(\omega+1)+(\omega+1)$: simply take the first two copies of $\omega+1$ and move then to the end.
So by contraposition: if $X$ does not contain $\lambda+\omega+\omega$ then it also does not contain $\lambda+\omega$, so that $\lambda+\omega+\omega$ is never `the smallest'. 
This does suggest, however, a modification of the question: look at indecomposable $\lambda$s.