Sum and sum only.

$\begin{aligned} S & = \sum_{n=1}^{10} n \left(\frac {1^2}{1+n} + \frac {2^2}{2+n} + \frac {3^2}{3+n} + \cdots + \frac {10^2}{10+n} \right) \\ & = \sum_{n=1}^{10} \sum_{m=1}^{10} \frac {m^2n}{m+n} \quad \quad \small \color{#3D99F6} \text{Note that }\sum_{n=1}^{10} \sum_{m=1}^{10} \frac {m^2n}{m+n} = \sum_{n=1}^{10} \sum_{m=1}^{10} \frac {n^2m}{m+n} \\ \implies 2S & = \sum_{n=1}^{10} \sum_{m=1}^{10} \frac {m^2n}{m+n} + \sum_{n=1}^{10} \sum_{m=1}^{10} \frac {n^2m}{m+n} \\ \implies S & = \frac 12 \sum_{n=1}^{10} \sum_{m=1}^{10} \frac {m^2n+n^2m}{m+n} \\ & = \frac 12 \sum_{n=1}^{10} \sum_{m=1}^{10} mn \\ & = \frac 12 \sum_{n=1}^{10} n \sum_{m=1}^{10} m \\ & = \frac 12 \times \frac {10(11)}2 \times \frac {10(11)}2 \\ & = \frac {55^2}2 \end{aligned}$

Therefore, $a+b = 55+2 = \boxed{57}$ .

Mathematica

Sum[n*Sum[k^2/(k + n), {k, 10}], {n, 10}]

3025/2

@Chew-Seong Cheong , sir please please post the solution. i have a bit of doubts in this one.