Direct form I and Direct form II
no.  but, for a causal LTI digital filter with a rational transfer function, it is necessary that the number of zeros, $M$, not exceed the number of poles, $N$.

$$\begin{align}
\\
 H(z) \triangleq \frac{Y(z)}{X(z)} &amp; = H_0 \frac{(z-q_1)(z-q_2)(z-q_3)...(z-q_M)}{(z-p_1)(z-p_2)(z-p_3)...(z-p_N)} \\
\\
 &amp; = H_0 \frac{z^M (1-q_1 z^{-1})(1-q_2 z^{-1})(1-q_3 z^{-1})...(1-q_M z^{-1})}{z^N (1-p_1 z^{-1})(1-p_2 z^{-1})(1-p_3 z^{-1})...(1-p_N z^{-1})}  \\
\\
 &amp; = z^{M-N} \ \frac{H_0 \ (1-q_1 z^{-1})(1-q_2 z^{-1})(1-q_3 z^{-1})...(1-q_M z^{-1})}{(1-p_1 z^{-1})(1-p_2 z^{-1})(1-p_3 z^{-1})...(1-p_N z^{-1})}  \\
\\
 &amp; = z^{M-N} \ \frac{b_0 + b_1 z^{-1} + b_2 z^{-2} + b_3 z^{-3} + ... + b_M z^{-M} }{ 1 + a_1 z^{-1} + a_2 z^{-2} + a_3 z^{-3} + ... +  + a_N z^{-N} }
\end{align}$$

this gets manipulated:

$$\begin{align} Y(z) &amp; \left(  1 + a_1 z^{-1} + a_2 z^{-2} + ... + a_N z^{-N}  \right) \\ 
&amp; \quad = X(z) z^{M-N} \left( b_0 + b_1 z^{-1} + b_2 z^{-2} + ... + b_M z^{-M} \right) \end{align}$$

or

$$\begin{align} Y(z) = \ &amp; b_0 z^{M-N} X(z) + b_1 z^{M-N-1} X(z) + b_2 z^{M-N-2} X(z) + ...  \\
 &amp; \quad ... + b_M z^{-N} X(z) - a_1 z^{-1}Y(z) - a_2 z^{-2}Y(z) - ... - a_N z^{-N}Y(z) \end{align}$$

inverting the Z Transform, this becomes 

$$\begin{align} y[n] = \ &amp; b_0 \ x[n+M-N] + b_1 \ x[n+M-N-1] + b_2 \ x[n+M-N-2] + ...  \\
 &amp; \quad ... + b_M \ x[n-N] - a_1 \ y[n-1] - a_2 \ y[n-2] - ... - a_N \ y[n-N] \end{align}$$

for a causal system or filter, we cannot look into the future for any input sample $x[n]$.  that means $M \le N$ so that there is no negative delay on the $x[n+M-N]$ term.  a negative delay means the LTI filter is peeking into the future, which is not causal.

but if you remove that leading delay factor and have this:

$$ H(z) = \frac{b_0 + b_1 z^{-1} + b_2 z^{-2} + b_3 z^{-3} + ... + b_M z^{-M} }{ 1 + a_1 z^{-1} + a_2 z^{-2} + a_3 z^{-3} + ... +  + a_N z^{-N} } $$

there is no reason that $M$ cannot be larger than $N$. but the number of poles will still be as large as $M$. if $M&gt;N$, then some of the poles will have to be at the origin $z=0$. in fact, if $N=0$, then you have an FIR filter and all of the poles are at the origin.