Obviously we have a,b,c,d >0. WLOG, let $ a\ge b\ge c\ge d\ge 0 $ So $ (b+c+d)^{2010}\ge (a+c+d)^{2010}\ge (a+b+d)^{2010}\ge (a+b+c)^{2010} $, which means $ (b+c+d)\ge (a+c+d)\ge (a+b+d)\ge (a+b+c) $, so $ d\ge c\ge b\ge a $, this implies that $ a=b=c=d $, Now we have this: $ (3a)^{2010}=3a $, so the only answer to the problem is $ a = b = c = d =\frac{1}{3} $

$a=b=c=d=0$ is self-evident.

$\ b-a=3^{\frac{1}{2010}}(a^{\frac{1}{2010}}-b^{\frac{1}{2010}})$ see if a is a solution then as $ \ 3a=(b+c+d)^{2010}$ so $\ a \geq 0 $ so $\ a^{\frac{1}{2010}}$ is well defined and nonnegative(the negative value can't occur since then b+c+d <0 which means atleast one of b,c,d is <0 which can't happen) . then see b>a or b<a gives contradiction. so b=a.similarly b=c and c=d. so a=b=c=d are cases to consider. if a=0 that gives solution. if a is non zero then a=1/3. so solutions are $\ a=b=c=d=0$ and $\ a=b=c=d=\frac{1}{3} $

We can say that $d$ is the greatest one and they are all greater than $0$. So, let $d \geq c \geq b \geq a \geq 0$. Now, using that, we can say that $b+c+d \geq a+c+d \geq a+b+d \geq a+b+c$ and $3a=(b+c+d)^{2010}\geq (a+c+d)^{2010}\geq (a+b+d)^{2010}\geq (a+b+c)^{2010}$, so now is $a \geq b \geq c \geq d$ $\implies$ $a=b=c=d$. So, $(3b)^{2010}=3b$ or $3^{2009}b^{2010}=b$ and finally $a=b=c=d=0$ or $a=b=c=d=\frac{1}{3}$

$k = a + b + c + d,p = b +c + d, q = a + c + d, r = a+b+d, s = a+b+c.$ $min(a,b,c,d)\geq 0 \implies min(p,q,r,s)\geq 0.$ $f(x)=x^{2010}-3(k - x)$, then $f'(x)=2010x^{2019}+3>0 $ (for $x \geq 0$) $\implies f(x)=0$ has at most 1 non-negative solution for fixed $k$. $f(p)=f(q)=f(r)=f(s)\implies p = q = r = s \implies k = \frac{4}{3} p\implies p^{2010}=p \implies p \in \{0, 1\} \implies a=b=c=d=0$ or $a=b=c=d=\frac{1}{3}$.

Answer. We claim that the only solutions are $(0,0,0,0)$ and $(\frac{1}{3},\frac{1}{3},\frac{1}{3},\frac{1}{3})$. Proof. Observe that the LHS is non-negative everywhere, so the RHS is also non-negative, thus $a,b,c,d\geq 0$. Suppose that not all are equal, then WLOG $a>b$. Then $3a=(b+c+d)^{2010}>3b=(a+c+d)^{2010}$. Since $b+c+d\geq 0$, $a+c+d \geq 0$, latter implies $b+c+d>a+c+d \iff b>a$ which is a clear contradiction. Then all $a,b,c,d$ are equal and we neet to solve $3a=(3a)^{2010}$. Let $x=3a$. Then we need to solve $x=x^{2010} \iff x(x^{2009}-1)=0$. One solution is $x=0$, the other is $1$ and there are no others since $f(x)=x^{2009}$ is strictly increasing. Thus the only solutions are $(0,0,0,0)$ and $(\frac{1}{3},\frac{1}{3},\frac{1}{3},\frac{1}{3})$, as desired. $\blacksquare$