Pick a big number N. The number of 5-tuples (x_1,x_2,...,x_5), such that x_1^3+x_2^5+...+x_5^11 is smaller or equal to N is not more than f(N^(1/3))*f(N^(1/5))*...*f(N^(1/11))<=N^(1/3+1/5+...+1/11)< N^0.9 (where f is floor function), but N/N^0.9=N^0.1, which tends to infinty for big N, so there are infinitely many numbers which are not representable as this power sum.

Let's assume for the sake of contradiction the set \[ S:=\left\{n \in \mathbb{N}:n=\sum_{i=1}^5{x_i^{2i+1}}\text{ for some }x_i \in \mathbb{N}\right\} \] is not finite. Then $|\mathbb{N} \setminus S|=n_0$ for some $n_0 \in \mathbb{N}$. But for all reals $x>0$ and positive integers $h$ we have \[ | \{n^h:n \in \mathbb{N}\}\cap [0,x] | \le (x+1)^{h^{-1}} = \mathcal{O}(x^{h^{-1}}) \] It means that, given positive integer $h_1,h_2,\ldots,h_m$, in the best case we'll have: \[ \left|\bigcup_{i=1}^m{\{n^{h_i}:n \in \mathbb{N}\}} \cap [0,x]\right|=\mathcal{O}(x^{\sum_{i=1}^m{h_i^{-1}}}). \] But in our case $\sum_{i=1}^5{(2i+1)^{-1}}<1$, so it implies that \[ |S\cap [0,x]|=\mathcal{O}(x^{\sum_{i=1}^5{(2i+1)^{-1}}})=o(x) \]. In other words $\mathbb{N} \setminus S$ cannot be finite, and that's a contradiction.

Any better answer??

AndrejK wrote: Pick a big number N. The number of 5-tuples (x_1,x_2,...,x_5), such that x_1^3+x_2^5+...+x_5^11 is smaller or equal to N is not more than f(N^(1/3))*f(N^(1/5))*...*f(N^(1/11))<=N^(1/3+1/5+...+1/11)< N^0.9 (where f is floor function), but N/N^0.9=N^0.1, which tends to infinty for big N, so there are infinitely many numbers which are not representable as this power sum. This solution is good but let me rewrite it better: Start counting number of numbers which are less than N and they can be written in this form:($=K$) $$K =\sqrt[3]{N}\times\sqrt[5]{N}\times\sqrt[7]{N}\times\sqrt[9]{N}\times\sqrt[11]{N}$$$$=N^{\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}}\approx N^{0.87}$$Now pick N a big number like $N=x^{100}$ $\Rightarrow$number of numbers which can not be written in that form=$N-K =x^{100}-x^{87}$ And this polynomial ( $x^{100}-x^{87}$ ) for big $N$s $\rightarrow \infty$ which means there are infinite N