$$\begin{array}{ll} \text{maximize} & \frac{a}{100} + \frac{4b}{100} + \frac{9c}{100} + \frac{16d}{100} + \frac{25e}{100} + \frac{36f}{100} + \frac{49g}{100} + \frac{64h}{100} + \frac{81i}{100} + \frac{100j}{100}\\ \text{subject to} & a+b+c+d+e+f+g+h+i+j = 100\end{array}$$
When I use the Lagrange multiplier and I take all my partial derivatives, I do not get any of the variables in terms of $\lambda$. For example, when I take the derivative with respect to $a$, I get $$\frac{1}{100} = \lambda$$ and with respect to $b$, I get $$\frac{4}{100} = \lambda$$
Is there any way I can find the values of $a$, $b$, $c$, $d$, $e$, $f$, $g$, $h$, $i$ and $j$ which maximises my function?
You get no results with Lagrange multipliers because there is no maximum. Take any point $(a,b,\ldots,j)$ which fulfills the constraint. Then $(a-1,b,c,\ldots,i,j+1)$ also fulfills the constraint, and the function value is greater there.
If you require that all variables are non-negative, then there is a maximum at $a=b=\cdots=i=0, j=100$.