= Solution
If the possible values have <probabilities> $p_1,\ldots,p_N$, allowing zeros, the <Shannon entropy> in bits is $H(X)=-\sum_i p_i\log_2p_i$, with $0\log_20=0$. Each term is nonnegative, so $H\geq0$, with equality precisely when one value has <probability> one.
For the upper bound, prove first that $\log t\leq t-1$ for $t>0$: the function $t-1-\log t$ has derivative $1-1/t$ and its unique minimum, zero, at $t=1$. Let $m$ be the number of positive <probabilities>. Applying this inequality to $t=(Np_i)^{-1}$ gives
$$
\sum_{p_i>0}p_i\log\frac1{Np_i}\leq\sum_{p_i>0}\left(\frac1N-p_i\right)=\frac mN-1\leq0.
$$
The left side is $H(X)\log2-\log N$, proving
$$
\boxed{0\leq H(X)\leq\log_2N.}
$$
Equality at the maximum requires $m=N$ and equality in every logarithmic inequality, so every $p_i=1/N$. Conversely the <uniform distribution> attains it. For $N=1$ the maximum and minimum coincide at zero.
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